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Three Dimensional Geometry — JEE Previous Year Questions

Every Three Dimensional Geometry question asked in JEE Main and JEE Advanced across the last 186 papers — 344 questions, each with its correct answer. Free to read, no account needed.

Questions

344

Papers it appeared in

176/186

Appearance rate

95%

All 344 Three Dimensional Geometry questions

Most recent papers first.

Q1·MathematicsMultiple correctJEE Advanced 2026
Let PPP be the plane such that it contains the straight line x−12=y−33=z+21\frac{x-1}{2} = \frac{y-3}{3} = \frac{z+2}{1}2x−1​=3y−3​=1z+2​ and is perpendicular to the plane x+2y+3z=4x + 2y + 3z = 4x+2y+3z=4. Let P1P_1P1​ be the plane which passes through the point (4,2,2)(4, 2, 2)(4,2,2) and is parallel to PPP. Then which of the following statements is (are) TRUE ?
  1. (A)The equation of the plane PPP is 7x−5y+z=−107x - 5y + z = -107x−5y+z=−10
  2. (B)The distance between the planes PPP and P1P_1P1​ is 30
  3. (C)The distance of the plane PPP from the origin is 232\sqrt{3}23​
  4. (D)The acute angle between the plane PPP and the plane 2x+2y+z=32x + 2y + z = 32x+2y+z=3 is cos⁡−1(133)\cos^{-1}\left( \frac{1}{3\sqrt{3}} \right)cos−1(33​1​)

Correct answer: (A), (D)

Step-by-step solution →
Q2·MathematicsMultiple correctJEE Advanced 2026
Let LLL be the straight line joining the points P(1,2,−1)P(1, 2, -1)P(1,2,−1) and Q(2,3,1)Q(2, 3, 1)Q(2,3,1). Let SSS be the foot of the perpendicular drawn from the point R(4,−1,5)R(4, -1, 5)R(4,−1,5) to the line LLL. Another line passing through RRR intersects LLL at a point TTT such that the point SSS divides the line segment PTPTPT internally in the ratio ∣PS∣:∣ST∣=1:2|PS| : |ST| = 1 : 2∣PS∣:∣ST∣=1:2, where ∣PS∣|PS|∣PS∣ and ∣ST∣|ST|∣ST∣ are the lengths of the line segments PSPSPS and STSTST, respectively. Then which of the following statements is (are) TRUE ?
  1. (A)The orthocentre of the triangle PRTPRTPRT is (235,−4,315)\left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)(523​,−4,531​)
  2. (B)The orthocentre of the triangle PRTPRTPRT is (4,3,5)(4, 3, 5)(4,3,5)
  3. (C)The area of the triangle PRTPRTPRT is 656\sqrt{5}65​
  4. (D)The area of the triangle PRTPRTPRT is 18518\sqrt{5}185​

Correct answer: (A), (D)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let the foot of perpendicular from the point (λ,2,3)(\lambda, 2, 3)(λ,2,3) on the line x−41=y−92=z−51\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}1x−4​=2y−9​=1z−5​ be the point (1,μ,2)(1, \mu, 2)(1,μ,2). Then the distance between the lines x−12=y−23=z+46\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}2x−1​=3y−2​=6z+4​ and x−λ2=y−μ3=z+56\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}2x−λ​=3y−μ​=6z+5​ is equal to:
  1. (A)127\dfrac{12}{7}712​
  2. (B)1457\dfrac{\sqrt{145}}{7}7145​​
  3. (C)1467\dfrac{\sqrt{146}}{7}7146​​
  4. (D)1437\dfrac{\sqrt{143}}{7}7143​​

Correct answer: (C)

Step-by-step solution →
Q4·MathematicsNumericalJEE Main 2026
Let a line L1L_1L1​ pass through the origin and be perpendicular to the lines L2:r⃗=(3+t)i^+(2t−1)j^+(2t+4)k^L_2 : \vec{r} = (3 + t)\hat{i} + (2t - 1)\hat{j} + (2t + 4)\hat{k}L2​:r=(3+t)i^+(2t−1)j^​+(2t+4)k^ and L3:r⃗=(3+2s)i^+(3+2s)j^+(2+s)k^L_3 : \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}L3​:r=(3+2s)i^+(3+2s)j^​+(2+s)k^, t,s∈Rt, s \in \mathbb{R}t,s∈R. If (a,b,c)(a, b, c)(a,b,c), a∈Za \in \mathbb{Z}a∈Z, is the point on L3L_3L3​ at a distance of 17\sqrt{17}17​ from the point of intersection of L1L_1L1​ and L2L_2L2​, then (a+b+c)2(a + b + c)^2(a+b+c)2 is equal to ________.

Correct answer: 4

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
The shortest distance between the lines x−41=y−32=z−2−3\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}1x−4​=2y−3​=−3z−2​ and x+22=y−64=z−5−5\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}2x+2​=4y−6​=−5z−5​ is :
  1. (A)566\frac{5\sqrt{6}}{6}656​​
  2. (B)252\sqrt{5}25​
  3. (C)353\sqrt{5}35​
  4. (D)454\sqrt{5}45​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsNumericalJEE Main 2026
Let the image of the point P(0,−5,0)P(0, -5, 0)P(0,−5,0) in the line x−12=y1=z+1−2\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}2x−1​=1y​=−2z+1​ be the point RRR and the image of the point Q(0,−12,0)Q\left(0, \frac{-1}{2}, 0\right)Q(0,2−1​,0) in the line x−1−1=y+94=z+11\frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1}−1x−1​=4y+9​=1z+1​ be the point SSS. Then the square of the area of the parallelogram PQRSPQRSPQRS is __________.

Correct answer: 162

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let the image of the point P(1, 6, a) in the line L : x1=y−12=z−a+1b\frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}1x​=2y−1​=bz−a+1​, b > 0, be (a3,0,a+c)\left(\frac{a}{3}, 0, a+c\right)(3a​,0,a+c). If S(α, β, γ), α > 0, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is 2142\sqrt{14}214​, then α + β + γ is equal to:
  1. (A)19
  2. (B)20
  3. (C)21
  4. (D)22

Correct answer: (C)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let a line L be perpendicular to both the lines L1_11​ : x+13=y+35=z+57\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}3x+1​=5y+3​=7z+5​ and L2_22​ : x−21=y−44=z−67\frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}1x−2​=4y−4​=7z−6​. If θ is the acute angle between the lines L and L3_33​ : x−872=y−471=z2\frac{x-\frac{8}{7}}{2} = \frac{y-\frac{4}{7}}{1} = \frac{z}{2}2x−78​​=1y−74​​=2z​, then tan θ is equal to:
  1. (A)322\frac{3}{2}\sqrt{2}23​2​
  2. (B)522\frac{5}{2}\sqrt{2}25​2​
  3. (C)532\frac{5}{3}\sqrt{2}35​2​
  4. (D)432\frac{4}{3}\sqrt{2}34​2​

Correct answer: (B)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
The square of the distance of the point P(5,6,7)P(5,6,7)P(5,6,7) from the line x−22=y−53=z−24\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}2x−2​=3y−5​=4z−2​ is equal to:
  1. (A)3
  2. (B)5
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
The square of the distance of the point of intersection of the lines r⃗=(i^+j^−k^)+λ(ai^−j^)\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a\hat{i} - \hat{j})r=(i^+j^​−k^)+λ(ai^−j^​), a≠0a \neq 0a=0 and r⃗=(4i^−k^)+μ(2i^+ak^)\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + a\hat{k})r=(4i^−k^)+μ(2i^+ak^) from the origin is:
  1. (A)5
  2. (B)10
  3. (C)17
  4. (D)26

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let a triangle PQR be such that P and Q lie on the line x+38=y−42=z+12\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}8x+3​=2y−4​=2z+1​ and are at a distance of 6 units from R(1, 2, 3). If (α, β, γ) is the centroid of △PQR, then α + β + γ is equal to :
  1. (A)4
  2. (B)5
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line x1=y−11=z−22\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}1x​=1y−1​=2z−2​ is 4, then the sum of all possible values of a is equal to :
  1. (A)11
  2. (B)9
  3. (C)6
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
A line with direction ratios 1,−1,21, -1, 21,−1,2 intersects the lines x2=y3=z+13\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}2x​=3y​=3z+1​ and x+1−1=y−21=z4\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}−1x+1​=1y−2​=4z​ at the points PPP and QQQ, respectively. If the length of the line segment PQPQPQ is α\alphaα, then 225α2225\alpha^2225α2 is equal to:
  1. (A)102410241024
  2. (B)101410141014
  3. (C)110411041104
  4. (D)120412041204

Correct answer: (B)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
The shortest distance between the lines r⃗=(13i^+2j^+83k^)+λ(2i^−5j^+6k^)\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})r=(31​i^+2j^​+38​k^)+λ(2i^−5j^​+6k^) and r⃗=(−23i^−13k^)+μ(j^−k^)\vec{r} = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) + \mu(\hat{j} - \hat{k})r=(−32​i^−31​k^)+μ(j^​−k^), λ,μ∈R\lambda, \mu \in \mathbb{R}λ,μ∈R, is:
  1. (A)5\sqrt{5}5​
  2. (B)3
  3. (C)232\sqrt{3}23​
  4. (D)15\sqrt{15}15​

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
If (2α+1,α2−3α,α−12)\left(2\alpha + 1, \alpha^{2} - 3\alpha, \frac{\alpha - 1}{2}\right)(2α+1,α2−3α,2α−1​) is the image of (α,2α,1)(\alpha, 2\alpha, 1)(α,2α,1) in the line x−23=y−12=z1\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}3x−2​=2y−1​=1z​, then the possible value(s) of α is (are)
  1. (A)Only 3
  2. (B)Only 3 and −1
  3. (C)Only 3, 14\frac{1}{4}41​ and −1
  4. (D)Only 3 and 14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
The square of the distance of the point (−2,−8,6)(-2, -8, 6)(−2,−8,6) from the line x−11=y−12=z−1\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}1x−1​=2y−1​=−1z​ along the line x+51=y+5−1=z2\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}1x+5​=−1y+5​=2z​ is equal to:
  1. (A)333
  2. (B)666
  3. (C)888
  4. (D)121212

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let a line LLL passing through the point (1,1,1)(1, 1, 1)(1,1,1) be perpendicular to both the vectors 2i^+2j^+k^2\hat{i} + 2\hat{j} + \hat{k}2i^+2j^​+k^ and i^+2j^+2k^\hat{i} + 2\hat{j} + 2\hat{k}i^+2j^​+2k^. If P(a,b,c)P(a, b, c)P(a,b,c) is the foot of perpendicular from the origin on the line LLL, then the value of 34(a+b+c)34(a + b + c)34(a+b+c) is :
  1. (A)50
  2. (B)80
  3. (C)100
  4. (D)120

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
Let the point A be the foot of perpendicular drawn from the point P(a,b,0)P(a, b, 0)P(a,b,0) on the line x−12=y−21=z−α3\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}2x−1​=1y−2​=3z−α​. If the midpoint of the line segment PA is (0,34,−14)\left(0, \frac{3}{4}, \frac{-1}{4}\right)(0,43​,4−1​), then the value of a2+b2+α2a^2 + b^2 + \alpha^2a2+b2+α2 is equal to:
  1. (A)111
  2. (B)222
  3. (C)666
  4. (D)999

Correct answer: (A)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
If the point of intersection of the lines x+13=y+a5=z+b+17\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}3x+1​=5y+a​=7z+b+1​ and x−21=y−b4=z−2a7\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}1x−2​=4y−b​=7z−2a​ lies on xy-plane, then the value of a+ba+ba+b is :
  1. (A)2
  2. (B)5
  3. (C)7
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q20·MathematicsNumericalJEE Main 2026
If the distance of the point P(43, α\alphaα, β\betaβ), β<0\beta < 0β<0, from the line r⃗=4i^−k^+μ(2i^+3k^)\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k})r=4i^−k^+μ(2i^+3k^), μ∈R\mu \in \mathbf{R}μ∈R along a line with direction ratios 3, –1, 0 is 131013\sqrt{10}1310​, then α2+β2\alpha^2 + \beta^2α2+β2 is equal to ________.

Correct answer: 170

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2026
Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line x−11=y2=z−11\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}1x−1​=2y​=1z−1​. Then the distance of Q from the line x−93=y−92=z−5−2\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}3x−9​=2y−9​=−2z−5​ is
  1. (A)6
  2. (B)8
  3. (C)7
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
If the distances of the point (1, 2, a) from the line x−11=y2=z−11\dfrac{x-1}{1} = \dfrac{y}{2} = \dfrac{z-1}{1}1x−1​=2y​=1z−1​ along the lines L1:x−13=y−24=z−abL_1 : \dfrac{x-1}{3} = \dfrac{y-2}{4} = \dfrac{z-a}{b}L1​:3x−1​=4y−2​=bz−a​ and L2:x−11=y−24=z−acL_2 : \dfrac{x-1}{1} = \dfrac{y-2}{4} = \dfrac{z-a}{c}L2​:1x−1​=4y−2​=cz−a​ are equal, then a + b + c is equal to
  1. (A)7
  2. (B)5
  3. (C)6
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsNumericalJEE Main 2026
Let a line L passing through the point P(1, 1, 1) be perpendicular to the lines x−44=y−11=z−11\frac{\mathrm{x} - 4}{4} = \frac{\mathrm{y} - 1}{1} = \frac{\mathrm{z} - 1}{1}4x−4​=1y−1​=1z−1​ and x−171=y−711=z0\frac{\mathrm{x} - 17}{1} = \frac{\mathrm{y} - 71}{1} = \frac{\mathrm{z}}{0}1x−17​=1y−71​=0z​. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point S(1,0, –1) intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.

Correct answer: 6

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2026
Let the lines L1L_{1}L1​: r⃗=i^+2j^+3k^+λ(2i^+3j^+4k^)\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda\left(2\hat{i}+3\hat{j}+4\hat{k}\right)r=i^+2j^​+3k^+λ(2i^+3j^​+4k^), λ∈ℝ and L2L_{2}L2​ : r⃗=(4i^+j^)+μ(5i^+2j^+k^)\vec{r}=\left(4\hat{i}+\hat{j}\right)+\mu\left(5\hat{i}+2\hat{j}+\hat{k}\right)r=(4i^+j^​)+μ(5i^+2j^​+k^), μ∈ℝ, intersect at the point R. Let P and Q be the points lying on lines L1L_{1}L1​ and L2L_{2}L2​, respectively, such that ∣PR→∣=29\left|\overrightarrow{PR}\right|=\sqrt{29}​PR​=29​ and ∣PQ→∣=473\left|\overrightarrow{PQ}\right|=\sqrt{\frac{47}{3}}​PQ​​=347​​ . If the point P lies in the first octant, then 27(QR)227(QR)^{2}27(QR)2 is equal to
  1. (A)340340340
  2. (B)360360360
  3. (C)320320320
  4. (D)348348348

Correct answer: (B)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2026
The sum of all values of α\alphaα, for which the shortest distance between the lines x+1α=y−2−1=z−4−α\frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha}αx+1​=−1y−2​=−αz−4​ and xα=y−12=z−12α\frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha}αx​=2y−1​=2αz−1​ is 2\sqrt{2}2​, is
  1. (A)8
  2. (B)−6
  3. (C)6
  4. (D)−8

Correct answer: (B)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2026
Let the direction cosines of two lines satisfy the equations : 4ℓ+m−n=04\ell + m - n = 04ℓ+m−n=0 and 2mn+10nℓ+3ℓm=02mn + 10n\ell + 3\ell m = 02mn+10nℓ+3ℓm=0. Then the cosine of the acute angle between these lines is :
  1. (A)1038\frac{10}{\sqrt{38}}38​10​
  2. (B)20338\frac{20}{3\sqrt{38}}338​20​
  3. (C)10738\frac{10}{7\sqrt{38}}738​10​
  4. (D)10338\frac{10}{3\sqrt{38}}338​10​

Correct answer: (D)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2026
The vertices B and C of a triangle ABC lie on the line x1=1−y−2=z−23\frac{x}{1} = \frac{1-y}{-2} = \frac{z-2}{3}1x​=−21−y​=3z−2​. The coordinates of A and B are (1, 6, 3) and (4, 9, α\alphaα) respectively and C is at a distance of 10 units from B. The area (in sq. units) of ΔABC\Delta ABCΔABC is:
  1. (A)5135\sqrt{13}513​
  2. (B)151315\sqrt{13}1513​
  3. (C)201320\sqrt{13}2013​
  4. (D)101310\sqrt{13}1013​

Correct answer: (A)

Step-by-step solution →
Q28·MathematicsNumericalJEE Main 2026
If the image of the point P(a, 2, a) in the line x2=y+a1=z1\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}2x​=1y+a​=1z​ is Q and the of image of Q in the line x−2b2=y−a1=z+2b−5\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}2x−2b​=1y−a​=−5z+2b​ is P, then a + b is equal to………..

Correct answer: 3

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2026
Let P(α, β, γ) be the point on the line x−12=y+1−3=z\frac{x-1}{2} = \frac{y+1}{-3} = z2x−1​=−3y+1​=z at a distance 4144\sqrt{14}414​ from the point (1, −1, 0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}1x−α​=2y−β​=3z−γ​ and x+52=y−101=z−31\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}2x+5​=1y−10​=1z−3​ , is equal to
  1. (A)7547\sqrt{\frac{5}{4}}745​​
  2. (B)4754\sqrt{\frac{7}{5}}457​​
  3. (C)4574\sqrt{\frac{5}{7}}475​​
  4. (D)2742\sqrt{\frac{7}{4}}247​​

Correct answer: (B)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2026
Let L be the line x+12=y+13=z+36\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}2x+1​=3y+1​=6z+3​ and let S be the set of all points (a, b, c) on L, whose distance from the line x+12=y+13=z−90\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}2x+1​=3y+1​=0z−9​ along the line L is 7. Then ∑(a,b,c)∈S(a+b+c)\sum\limits_{(a,b,c) \in S} (a + b + c)(a,b,c)∈S∑​(a+b+c) is equal to :
  1. (A)34
  2. (B)28
  3. (C)40
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2026
If the image of the point P (1, 2 , a) in the line x−63=y−72=7−z2\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}3x−6​=2y−7​=27−z​ is Q(5, b, c), then a2+b2+c2a^2 + b^2 + c^2a2+b2+c2 is equal to
  1. (A)293
  2. (B)264
  3. (C)298
  4. (D)283

Correct answer: (C)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2026
Let the line L pass through the point (−3, 5, 2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2, r, 1) is 143\sqrt{\frac{14}{3}}314​​, then the sum of all possible values of r is :
  1. (A)12
  2. (B)16
  3. (C)6
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2026
Let (α, β, γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line r⃗=(−i^+3j^+k^)+λ(2i^+3j^−k^)\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})r=(−i^+3j^​+k^)+λ(2i^+3j^​−k^). Then the length of the projection of the vector αi^+βj^+γk^\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}αi^+βj^​+γk^ on the vector 6i^+2j^+3k^6\hat{i}+2\hat{j}+3\hat{k}6i^+2j^​+3k^ is :
  1. (A)157\frac{15}{7}715​
  2. (B)4
  3. (C)187\frac{18}{7}718​
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2026
Let the line L1L_1L1​ be parallel to the vector −3i^+2j^+4k^-3\hat{i} + 2\hat{j} + 4\hat{k}−3i^+2j^​+4k^ and pass through the point (2, 6, 7) and the line L2L_2L2​ be parallel to the vector 2i^+j^+3k^2\hat{i} + \hat{j} + 3\hat{k}2i^+j^​+3k^ and pass through the point (4, 3, 5). If the line L3L_3L3​ is parallel to the vector −3i^+5j^+16k^-3\hat{i} + 5\hat{j} + 16\hat{k}−3i^+5j^​+16k^ and intersects the lines L1L_1L1​ and L2L_2L2​ at the points C and D, respectively, then ∣CD→∣2\left|\overrightarrow{CD}\right|^2​CD​2 is equal to :
  1. (A)171
  2. (B)290
  3. (C)312
  4. (D)89

Correct answer: (B)

Step-by-step solution →
Q35·MathematicsMultiple correctJEE Advanced 2025
Let L1L_1L1​ be the line of intersection of the planes given by the equations 2x+3y+z=42x + 3y + z = 42x+3y+z=4 and x+2y+z=5x + 2y + z = 5x+2y+z=5. Let L2L_2L2​ be the line passing through the point P(2,−1,3)P(2, -1, 3)P(2,−1,3) and parallel to L1L_1L1​. Let M denote the plane given by the equation 2x+y−2z=62x + y - 2z = 62x+y−2z=6. Suppose that the line L2L_2L2​ meets the plane M at the point Q. Let R be the foot of the perpendicular drawn from P to the plane M. The which of the following statements is (are) TRUE?
  1. (A)The length of the line segment PQ is 939\sqrt{3}93​
  2. (B)The length of the line segment QR is 15
  3. (C)The area of ΔPQR is 32234\frac{3}{2}\sqrt{234}23​234​
  4. (D)The acute angle between the line segments PQ and PR is cos⁡−1(123)\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)cos−1(23​1​)

Correct answer: (A), (C)

Step-by-step solution →
Q36·MathematicsIntegerJEE Main 2025
Let the area of the triangle formed by the lines x+2=y−1=zx+2=y-1=zx+2=y−1=z, x−35=y−1=z−11\dfrac{x-3}{5}=\dfrac{y}{-1}=\dfrac{z-1}{1}5x−3​=−1y​=1z−1​ and x−3=y−33=z−21\dfrac{x}{-3}=\dfrac{y-3}{3}=\dfrac{z-2}{1}−3x​=3y−3​=1z−2​ be AAA. Then A2A^2A2 is equal to ___

Correct answer: 56

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let the values of λ\lambdaλ for which the shortest distance between the lines x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}2x−1​=3y−2​=4z−3​ and x−λ3=y−44=z−55\dfrac{x-\lambda}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}3x−λ​=4y−4​=5z−5​ is 16\dfrac{1}{\sqrt6}6​1​ be λ1\lambda_1λ1​ and λ2\lambda_2λ2​. Then the radius of the circle passing through the points (0,0)(0,0)(0,0), (λ1,λ2)(\lambda_1,\lambda_2)(λ1​,λ2​) and (λ2,λ1)(\lambda_2,\lambda_1)(λ2​,λ1​) is:
  1. (A)523\dfrac{5\sqrt2}{3}352​​
  2. (B)444
  3. (C)23\dfrac{\sqrt2}{3}32​​
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Let the line L pass through (1,1,1)(1,1,1)(1,1,1) and intersect the lines x−12=y+13=z−14\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}2x−1​=3y+1​=4z−1​ and x−31=y−42=z−11\dfrac{x-3}{1}=\dfrac{y-4}{2}=\dfrac{z-1}{1}1x−3​=2y−4​=1z−1​. Then, which of the following points lies on the line L?
  1. (A)(4,22,7)(4,22,7)(4,22,7)
  2. (B)(5,4,3)(5,4,3)(5,4,3)
  3. (C)(10,−29,−50)(10,-29,-50)(10,−29,−50)
  4. (D)(7,15,13)(7,15,13)(7,15,13)

Correct answer: (D)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
If the equation of the line passing through the point (0,−12,0)\left(0,-\dfrac{1}{2},0\right)(0,−21​,0) and perpendicular to the lines r⃗=λ(i^+aj^+bk^)\vec{r}=\lambda(\hat{i}+a\hat{j}+b\hat{k})r=λ(i^+aj^​+bk^) and r⃗=(i^−j^−6k^)+μ(−bi^+aj^+5k^)\vec{r}=(\hat{i}-\hat{j}-6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k})r=(i^−j^​−6k^)+μ(−bi^+aj^​+5k^) is x−1−2=y+4d=z−c−4\dfrac{x-1}{-2}=\dfrac{y+4}{d}=\dfrac{z-c}{-4}−2x−1​=dy+4​=−4z−c​, then a+b+c+da+b+c+da+b+c+d is equal to:
  1. (A)10
  2. (B)14
  3. (C)13
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
If the shortest distance between the lines x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}2x−1​=3y−2​=4z−3​ and x1=yα=z−51\dfrac{x}{1}=\dfrac{y}{\alpha}=\dfrac{z-5}{1}1x​=αy​=1z−5​ is 56\dfrac{5}{\sqrt{6}}6​5​, then the sum of all possible values of α\alphaα is
  1. (A)32\dfrac{3}{2}23​
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)3
  4. (D)−3-3−3

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Consider the lines L1:x−1=y−2=zL_1:x-1=y-2=zL1​:x−1=y−2=z and L2:x−2=y=z−1L_2:x-2=y=z-1L2​:x−2=y=z−1. Let the feet of the perpendiculars from the point P(5,1,−3)P(5,1,-3)P(5,1,−3) on the lines L1L_1L1​ and L2L_2L2​ be QQQ and RRR respectively. If the area of the triangle PQRPQRPQR is AAA, then 4A24A^24A2 is equal to:
  1. (A)139
  2. (B)147
  3. (C)151
  4. (D)143

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Let AAA and BBB be two distinct points on the line L:x−63=y−72=z−7−2L:\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}L:3x−6​=2y−7​=−2z−7​. Both AAA and BBB are at a distance 2172\sqrt{17}217​ from the foot of perpendicular drawn from the point (1,2,3)(1,2,3)(1,2,3) on the line LLL. If OOO is the origin, then OA→⋅OB→\overrightarrow{OA}\cdot\overrightarrow{OB}OA⋅OB is equal to:
  1. (A)49
  2. (B)47
  3. (C)21
  4. (D)62

Correct answer: (B)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
Let the shortest distance between the lines x−33=y−α−1=z−31\dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}3x−3​=−1y−α​=1z−3​ and x+3−3=y+72=z−β4\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}−3x+3​=2y+7​=4z−β​ be 3303\sqrt{30}330​. Then the positive value of 5α+β5\alpha+\beta5α+β is:
  1. (A)42
  2. (B)46
  3. (C)48
  4. (D)40

Correct answer: (B)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
Let AAA be the point of intersection of the lines L1:x−71=y−50=z−3−1L_1:\dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}L1​:1x−7​=0y−5​=−1z−3​ and L2:x−13=y+34=z+75L_2:\dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}L2​:3x−1​=4y+3​=5z+7​. Let BBB and CCC be the points on the lines L1L_1L1​ and L2L_2L2​ respectively such that AB=AC=15AB=AC=\sqrt{15}AB=AC=15​. Then the square of the area of the triangle ABCABCABC is:
  1. (A)54
  2. (B)63
  3. (C)57
  4. (D)60

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
Let the values of ppp, for which the shortest distance between the lines x+13=y4=z5\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}3x+1​=4y​=5z​ and r⃗=(pi^+2j^+k^)+λ(2i^+3j^+4k^)\vec{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})r=(pi^+2j^​+k^)+λ(2i^+3j^​+4k^) is 16\dfrac{1}{\sqrt6}6​1​, be a,ba, ba,b, (a<b)(a<b)(a<b). Then the length of the latus rectum of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1 is:
  1. (A)9
  2. (B)32\dfrac{3}{2}23​
  3. (C)23\dfrac{2}{3}32​
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
The distance of the point (7,10,11)(7,10,11)(7,10,11) from the line x−41=y−40=z−23\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}1x−4​=0y−4​=3z−2​ along the line x−92=y−133=z−176\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}2x−9​=3y−13​=6z−17​ is:
  1. (A)18
  2. (B)14
  3. (C)12
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
Let a line passing through the point (4,1,0)(4,1,0)(4,1,0) intersect the line L1:x−12=y−23=z−34L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}L1​:2x−1​=3y−2​=4z−3​ at the point A(α,β,γ)A(\alpha,\beta,\gamma)A(α,β,γ) and the line L2:x−6=y=−z+4L_2:x-6=y=-z+4L2​:x−6=y=−z+4 at the point B(a,b,c)B(a,b,c)B(a,b,c). Then ∣101αβγabc∣\begin{vmatrix}1&0&1\\\alpha&\beta&\gamma\\a&b&c\end{vmatrix}​1αa​0βb​1γc​​ is equal to:
  1. (A)8
  2. (B)16
  3. (C)12
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2025
Each of the angles β\betaβ and γ\gammaγ that a given line makes with the positive yyy- and zzz-axes, respectively, is half of the angle that this line makes with the positive xxx-axis. Then the sum of all possible values of the angle β\betaβ is:
  1. (A)3π4\dfrac{3\pi}{4}43π​
  2. (B)π\piπ
  3. (C)π2\dfrac{\pi}{2}2π​
  4. (D)3π2\dfrac{3\pi}{2}23π​

Correct answer: (A)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
Line L1L_1L1​ passes through the point (1,2,3)(1,2,3)(1,2,3) and is parallel to the z-axis. Line L2L_2L2​ passes through the point (λ,5,6)(\lambda,5,6)(λ,5,6) and is parallel to the y-axis. Let for λ=λ1,λ2\lambda=\lambda_1,\lambda_2λ=λ1​,λ2​ (λ2<λ1)(\lambda_2<\lambda_1)(λ2​<λ1​), the shortest distance between the two lines be 3. Then the square of the distance of the point (λ1,λ2,2)(\lambda_1,\lambda_2,2)(λ1​,λ2​,2) from the line L1L_1L1​ is:
  1. (A)40
  2. (B)32
  3. (C)25
  4. (D)37

Correct answer: (C)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2025
If the image of the point P(1,0,3)P(1, 0, 3)P(1,0,3) in the line joining the points A(4,7,1)A(4, 7, 1)A(4,7,1) and B(3,5,3)B(3, 5, 3)B(3,5,3) is Q(α,β,γ)Q(\alpha, \beta, \gamma)Q(α,β,γ), then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to:
  1. (A)473\dfrac{47}{3}347​
  2. (B)463\dfrac{46}{3}346​
  3. (C)18
  4. (D)13

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2025
Let the vertices QQQ and RRR of the triangle PQRPQRPQR lie on the line x+35=y−12=z+43\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}5x+3​=2y−1​=3z+4​, QR=5QR=5QR=5 and the coordinates of the point PPP be (0,2,3)(0,2,3)(0,2,3). If the area of the triangle PQRPQRPQR is mn\dfrac{m}{n}nm​ then:
  1. (A)m−521 n=0m-5\sqrt{21}\,n=0m−521​n=0
  2. (B)2m−521 n=02m-5\sqrt{21}\,n=02m−521​n=0
  3. (C)5m−221 n=05m-2\sqrt{21}\,n=05m−221​n=0
  4. (D)5m−212 n=05m-21\sqrt{2}\,n=05m−212​n=0

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2025
The line L1L_1L1​ is parallel to the vector a⃗=−3i^+2j^+4k^\vec{a}=-3\hat{i}+2\hat{j}+4\hat{k}a=−3i^+2j^​+4k^ and passes through the point (7,6,2)(7,6,2)(7,6,2) and the line L2L_2L2​ is parallel to the vector b⃗=2i^+j^+3k^\vec{b}=2\hat{i}+\hat{j}+3\hat{k}b=2i^+j^​+3k^ and passes through the point (5,3,4)(5,3,4)(5,3,4). The shortest distance between the lines L1L_1L1​ and L2L_2L2​ is:
  1. (A)2338\dfrac{23}{\sqrt{38}}38​23​
  2. (B)2157\dfrac{21}{\sqrt{57}}57​21​
  3. (C)2357\dfrac{23}{\sqrt{57}}57​23​
  4. (D)2138\dfrac{21}{\sqrt{38}}38​21​

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2025
Let a straight line L pass through the point P(2,−1,3)P(2,-1,3)P(2,−1,3) and be perpendicular to the lines x−12=y+11=z−3−2\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-2}2x−1​=1y+1​=−2z−3​ and x−31=y−23=z+24\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{4}1x−3​=3y−2​=4z+2​. If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is:
  1. (A)2
  2. (B)10\sqrt{10}10​
  3. (C)3
  4. (D)232\sqrt{3}23​

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2025
Let P be the foot of the perpendicular from the point (1,2,2)(1, 2, 2)(1,2,2) on the line L:x−11=y+1−1=z−22L:\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z-2}{2}L:1x−1​=−1y+1​=2z−2​. Let the line r⃗=(−i^−2k^)+λ(i^−j^+k^)\vec{r}=\left(-\hat{i}-2\hat{k}\right)+\lambda\left(\hat{i}-\hat{j}+\hat{k}\right)r=(−i^−2k^)+λ(i^−j^​+k^), λ∈R\lambda\in Rλ∈R, intersect the line L at Q. Then 2(PQ)22(PQ)^22(PQ)2 is equal to:
  1. (A)27
  2. (B)25
  3. (C)29
  4. (D)19

Correct answer: (A)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2025
Let L1:x−11=y−2−1=z−12L_1:\dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z-1}{2}L1​:1x−1​=−1y−2​=2z−1​ and L2:x+1−1=y−22=z1L_2:\dfrac{x+1}{-1}=\dfrac{y-2}{2}=\dfrac{z}{1}L2​:−1x+1​=2y−2​=1z​ be two lines. Let L3L_3L3​ be a line passing through the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and be perpendicular to both L1L_1L1​ and L2L_2L2​. If L3L_3L3​ intersects L1L_1L1​, then ∣5α−11β−8γ∣|5\alpha-11\beta-8\gamma|∣5α−11β−8γ∣ equals:
  1. (A)18
  2. (B)16
  3. (C)25
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2025
If the image of the point (4,4,3)(4, 4, 3)(4,4,3) in the line x−12=y−21=z−13\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}2x−1​=1y−2​=3z−1​ is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to
  1. (A)9
  2. (B)12
  3. (C)8
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2025
Let A(x,y,z)A(x, y, z)A(x,y,z) be a point in xy-plane, which is equidistant from three points (0,3,2)(0, 3, 2)(0,3,2), (2,0,3)(2, 0, 3)(2,0,3) and (0,0,1)(0, 0, 1)(0,0,1). Let B=(1,4,−1)B=(1, 4, -1)B=(1,4,−1) and C=(2,0,−2)C=(2, 0, -2)C=(2,0,−2). Then among the statements (S1): △ABC\triangle ABC△ABC is an isosceles right angled triangle and (S2): the area of △ABC\triangle ABC△ABC is 922\frac{9\sqrt{2}}{2}292​​.
  1. (A)both are true
  2. (B)only (S1) is true
  3. (C)only (S2) is true
  4. (D)both are false

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2025
The square of the distance of the point (157,327,7)\left(\frac{15}{7},\frac{32}{7},7\right)(715​,732​,7) from the line x+13=y+35=z+57\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^+7k^\hat{i}+4\hat{j}+7\hat{k}i^+4j^​+7k^ is:
  1. (A)54
  2. (B)41
  3. (C)66
  4. (D)44

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsIntegerJEE Main 2025
Let P be the image of the point Q(7,−2,5)Q(7,-2,5)Q(7,−2,5) in the line L:x−12=y+13=z4L:\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}L:2x−1​=3y+1​=4z​ and R(5,p,q)R(5,p,q)R(5,p,q) be a point on L. Then the square of the area of △PQR\triangle PQR△PQR is __________.

Correct answer: 957

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2025
Let the line passing through the points (−1,2,1)(-1,2,1)(−1,2,1) and parallel to the line x−12=y+13=z4\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}2x−1​=3y+1​=4z​ intersect the line x+23=y−32=z−41\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}3x+2​=2y−3​=1z−4​ at the point P. Then the distance of P from the point Q(4,−5,1)Q(4,-5,1)Q(4,−5,1) is :
  1. (A)555
  2. (B)101010
  3. (C)565\sqrt656​
  4. (D)555\sqrt555​

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2025
Let in a △ABC\triangle ABC△ABC, the length of the side AC be 6, the vertex B be (1,2,3)(1,2,3)(1,2,3) and the vertices A, C lie on the line x−63=y−72=z−7−2\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}3x−6​=2y−7​=−2z−7​. Then the area (in sq. units) of △ABC\triangle ABC△ABC is
  1. (A)424242
  2. (B)212121
  3. (C)565656
  4. (D)171717

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2025
If the square of the shortest distance between the lines x−21=y−12=z+3−3\dfrac{x-2}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{-3}1x−2​=2y−1​=−3z+3​ and x+12=y+34=z+5−5\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{-5}2x+1​=4y+3​=−5z+5​ is mn\dfrac{m}{n}nm​, where m,nm,nm,n are coprime numbers, then m+nm+nm+n is equal to :
  1. (A)6
  2. (B)9
  3. (C)21
  4. (D)14

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2025
The distance of the line x−22=y−63=z−34\dfrac{x-2}{2}=\dfrac{y-6}{3}=\dfrac{z-3}{4}2x−2​=3y−6​=4z−3​ from the point (1,4,0)(1,4,0)(1,4,0) along the line x1=y−22=z+33\dfrac{x}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{3}1x​=2y−2​=3z+3​ is :
  1. (A)17\sqrt{17}17​
  2. (B)14\sqrt{14}14​
  3. (C)15\sqrt{15}15​
  4. (D)13\sqrt{13}13​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2025
Let PPP be the foot of the perpendicular from the point Q(10,−3,−1)Q(10,-3,-1)Q(10,−3,−1) on the line x−37=y−2−1=z+1−2\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}7x−3​=−1y−2​=−2z+1​. Then the area of the right angled triangle PQRPQRPQR, where RRR is the point (3,−2,1)(3,-2,1)(3,−2,1), is
  1. (A)9159\sqrt{15}915​
  2. (B)30\sqrt{30}30​
  3. (C)8158\sqrt{15}815​
  4. (D)3303\sqrt{30}330​

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2025
The perpendicular distance, of the line x−12=y+2−1=z+32\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z+3}{2}2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1)P(2,-10,1)P(2,−10,1), is:
  1. (A)6
  2. (B)525\sqrt{2}52​
  3. (C)353\sqrt{5}35​
  4. (D)434\sqrt{3}43​

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2025
Let L1:x−12=y−23=z−34L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}L1​:2x−1​=3y−2​=4z−3​ and L2:x−23=y−44=z−55L_2:\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}L2​:3x−2​=4y−4​=5z−5​. Which point lies on the line of shortest distance between L1L_1L1​ and L2L_2L2​?
  1. (A)(−53,−7,1)\left(-\tfrac53,-7,1\right)(−35​,−7,1)
  2. (B)(2,3,13)\left(2,3,\tfrac13\right)(2,3,31​)
  3. (C)(83,−1,13)\left(\tfrac83,-1,\tfrac13\right)(38​,−1,31​)
  4. (D)(143,−3,223)\left(\tfrac{14}{3},-3,\tfrac{22}{3}\right)(314​,−3,322​)

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsIntegerJEE Main 2025
Let L1:x−13=y−1−1=z+10L_1:\dfrac{x-1}{3}=\dfrac{y-1}{-1}=\dfrac{z+1}{0}L1​:3x−1​=−1y−1​=0z+1​ and L2:x−22=y+40=z+4αL_2:\dfrac{x-2}{2}=\dfrac{y+4}{0}=\dfrac{z+4}{\alpha}L2​:2x−2​=0y+4​=αz+4​ meet at B. If P is the foot of perpendicular from A(1,−1,1)A(1,-1,1)A(1,−1,1) on L2L_2L2​, find 26 α (PB)26\,\alpha\,(PB)26α(PB) (as integer).

Correct answer: 216

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2025
Let a line pass through two distinct points P(−2,−1,3)P(-2,-1,3)P(−2,−1,3) and QQQ, and be parallel to the vector 3i^+2j^+2k^3\hat{i}+2\hat{j}+2\hat{k}3i^+2j^​+2k^. If the distance of the point QQQ from the point R(1,3,3)R(1,3,3)R(1,3,3) is 5, then the square of the area of △PQR\triangle PQR△PQR is equal to:
  1. (A)136
  2. (B)140
  3. (C)144
  4. (D)148

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsMultiple correctJEE Advanced 2024
A straight line drawn from the point P(1, 3, 2), parallel to the line x−21=y−42=z−61\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}1x−2​=2y−4​=1z−6​ intersects the plane L1:x−y+3z=6L_1 : x - y + 3z = 6L1​:x−y+3z=6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L1L_1L1​ intersects the plane L2:2x−y+z=−4L_2 : 2x - y + z = -4L2​:2x−y+z=−4 at the point R. then which of the following statements is(are) TRUE?
  1. (A)The length of the line segment PQ is 6\sqrt{6}6​
  2. (B)The coordinates of R are (1, 6, 3)
  3. (C)The centroid of the triangle PQR is (43,143,53)\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)(34​,314​,35​)
  4. (D)The perimeter of the triangle PQR is 2+6+11\sqrt{2} + \sqrt{6} + \sqrt{11}2​+6​+11​

Correct answer: (A), (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Advanced 2024
Let γ∈R\gamma \in Rγ∈R be such that the lines L1:x+111=y+212=z+293L_{1} : \frac{x + 11}{1} = \frac{y + 21}{2} = \frac{z + 29}{3}L1​:1x+11​=2y+21​=3z+29​ and L2:x+163=y+112=z+4γL_{2} : \frac{x + 16}{3} = \frac{y + 11}{2} = \frac{z + 4}{\gamma}L2​:3x+16​=2y+11​=γz+4​ intersect. Let R1R_{1}R1​ be the point of intersection of L1L_{1}L1​ and L2L_{2}L2​. Let O=(0,0,0)O = (0, 0, 0)O=(0,0,0), and n^\hat{n}n^ denote a unit normal vector to the plane containing both the lines L1L_{1}L1​ and L2L_{2}L2​. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.γ\gammaγ equals1.−i^−j^+k^-\hat{i} - \hat{j} + \hat{k}−i^−j^​+k^
Q.A possible choice for n^\hat{n}n^ is2.32\sqrt{\frac{3}{2}}23​​
R.OR1→\overrightarrow{OR_{1}}OR1​​ equals3.1
S.A possible value of OR1→⋅n^\overrightarrow{OR_{1}} \cdot \hat{n}OR1​​⋅n^ is4.16i^−26j^+16k^\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}6​1​i^−6​2​j^​+6​1​k^
5.23\sqrt{\frac{2}{3}}32​​
  1. (A)(P) →\to→ (3) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (2)
  2. (B)(P) →\to→ (5) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (2)
  3. (C)(P) →\to→ (3) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (5)
  4. (D)(P) →\to→ (3) (Q) →\to→ (1) (R) →\to→ (4) (S) →\to→ (5)

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsMultiple correctJEE Advanced 2024
Let R3R^{3}R3 denote the three dimensional space. Take two points P=(1,2,3)P = (1, 2, 3)P=(1,2,3) and Q=(4,2,7)Q = (4, 2, 7)Q=(4,2,7). Let dist(X,Y)\mathrm{dist}(X, Y)dist(X,Y) denote the distance between two points XXX and YYY in R3R^{3}R3. Let S={X∈R3:(dist(X,P))2−(dist(X,Q))2=50}S = \{X \in R^{3} : (\mathrm{dist}(X, P))^{2} - (\mathrm{dist}(X, Q))^{2} = 50\}S={X∈R3:(dist(X,P))2−(dist(X,Q))2=50} and T={Y∈R3:(dist(Y,Q))2−(dist(Y,P))2=50}T = \{Y \in R^{3} : (\mathrm{dist}(Y, Q))^{2} - (\mathrm{dist}(Y, P))^{2} = 50\}T={Y∈R3:(dist(Y,Q))2−(dist(Y,P))2=50}. Then which of the following statements is(are) TRUE ?
  1. (A)There is a triangle whose area is 1 and all of whose vertices are from SSS.
  2. (B)There are two distinct points LLL and MMM in TTT such that each point on the line segments LMLMLM is also in TTT.
  3. (C)There are infinitely many rectangles of perimeter 48, two of whose vertices are from SSS and the other two vertices are from TTT.
  4. (D)There is a square of perimeter 48, two of whose vertices are from SSS and the other two vertices are from TTT.

Correct answer: (A), (B), (C), (D)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2024
Consider the line L passing through the points (1,2,3)(1, 2, 3)(1,2,3) and (2,3,5)(2, 3, 5)(2,3,5). The distance of the point (113,113,193)\left(\dfrac{11}{3}, \dfrac{11}{3}, \dfrac{19}{3}\right)(311​,311​,319​) from the line L along the line 3x−112=3y−111=3z−192\dfrac{3x-11}{2}=\dfrac{3y-11}{1}=\dfrac{3z-19}{2}23x−11​=13y−11​=23z−19​ is equal to:
  1. (A)333
  2. (B)555
  3. (C)444
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correctJEE Main 2024
Let the line L intersect the lines x−2=−y=z−1x - 2 = -y = z - 1x−2=−y=z−1, 2(x+1)=2(y−1)=z+12(x + 1) = 2(y - 1) = z + 12(x+1)=2(y−1)=z+1 and be parallel to the line x−23=y−11=z−22\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}3x−2​=1y−1​=2z−2​. Then which of the following points lies on L?
  1. (A)(−13,1,1)\left(-\frac{1}{3}, 1, 1\right)(−31​,1,1)
  2. (B)(−13,1,−1)\left(-\frac{1}{3}, 1, -1\right)(−31​,1,−1)
  3. (C)(−13,−1,−1)\left(-\frac{1}{3}, -1, -1\right)(−31​,−1,−1)
  4. (D)(−13,−1,1)\left(-\frac{1}{3}, -1, 1\right)(−31​,−1,1)

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2024
The shortest distance between the line x−34=y+7−11=z−15\frac{x - 3}{4} = \frac{y + 7}{-11} = \frac{z - 1}{5}4x−3​=−11y+7​=5z−1​ and x−53=y−9−6=z+21\frac{x - 5}{3} = \frac{y - 9}{-6} = \frac{z + 2}{1}3x−5​=−6y−9​=1z+2​ is:
  1. (A)187563\frac{187}{\sqrt{563}}563​187​
  2. (B)178563\frac{178}{\sqrt{563}}563​178​
  3. (C)185563\frac{185}{\sqrt{563}}563​185​
  4. (D)179563\frac{179}{\sqrt{563}}563​179​

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsNumericalJEE Main 2024
The square of the distance of the image of the point (6,1,5)(6, 1, 5)(6,1,5) in the line x−13=y2=z−24\dfrac{x-1}{3}=\dfrac{y}{2}=\dfrac{z-2}{4}3x−1​=2y​=4z−2​, from the origin is _______.

Correct answer: 62

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2024
If the shortest distance between the lines L1:r⃗=(2+λ)i^+(1−3λ)j^+(3+4λ)k^L_1:\vec r=(2+\lambda)\hat i+(1-3\lambda)\hat j+(3+4\lambda)\hat kL1​:r=(2+λ)i^+(1−3λ)j^​+(3+4λ)k^, λ∈R\lambda\in\mathbb{R}λ∈R and L2:r⃗=2(1+μ)i^+3(1+μ)j^+(5+μ)k^L_2:\vec r=2(1+\mu)\hat i+3(1+\mu)\hat j+(5+\mu)\hat kL2​:r=2(1+μ)i^+3(1+μ)j^​+(5+μ)k^, μ∈R\mu\in\mathbb{R}μ∈R is mn\dfrac{m}{\sqrt n}n​m​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then the value of m+nm+nm+n equals:
  1. (A)384384384
  2. (B)387387387
  3. (C)377377377
  4. (D)390390390

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2024
If the shortest distance between the lines x−λ2=y−43=z−34\dfrac{x-\lambda}{2}=\dfrac{y-4}{3}=\dfrac{z-3}{4}2x−λ​=3y−4​=4z−3​ and x−24=y−46=z−78\dfrac{x-2}{4}=\dfrac{y-4}{6}=\dfrac{z-7}{8}4x−2​=6y−4​=8z−7​ is 1329\dfrac{13}{\sqrt{29}}29​13​, then a value of λ\lambdaλ is
  1. (A)−1325-\tfrac{13}{25}−2513​
  2. (B)1325\tfrac{13}{25}2513​
  3. (C)111
  4. (D)−1-1−1

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2024
Let P(x,y,z)P(x,y,z)P(x,y,z) be a point in the first octant, whose projection in the xy-plane is the point QQQ. Let OP=γOP=\gammaOP=γ; the angle between OQOQOQ and the positive x-axis be θ\thetaθ; and the angle between OPOPOP and the positive z-axis be ϕ\phiϕ, where OOO is the origin. Then the distance of PPP from the x-axis is:
  1. (A)γ1−sin⁡2ϕcos⁡2θ\gamma\sqrt{1-\sin^2\phi\cos^2\theta}γ1−sin2ϕcos2θ​
  2. (B)γ1+cos⁡2θsin⁡2ϕ\gamma\sqrt{1+\cos^2\theta\sin^2\phi}γ1+cos2θsin2ϕ​
  3. (C)γ1−sin⁡2θcos⁡2ϕ\gamma\sqrt{1-\sin^2\theta\cos^2\phi}γ1−sin2θcos2ϕ​
  4. (D)γ1+cos⁡2ϕsin⁡2θ\gamma\sqrt{1+\cos^2\phi\sin^2\theta}γ1+cos2ϕsin2θ​

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2024
Let P(α,β,γ)P(\alpha,\beta,\gamma)P(α,β,γ) be the image of the point Q(1,6,4)Q(1,6,4)Q(1,6,4) in the line x1=y−12=z−23\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}1x​=2y−1​=3z−2​. Then 2α+β+γ2\alpha+\beta+\gamma2α+β+γ is equal to _______ .

Correct answer: 11

Step-by-step solution →
Q80·MathematicsNumericalJEE Main 2024
If the shortest distance between the lines x−λ3=y−2−1=z−11\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}3x−λ​=−1y−2​=1z−1​ and x+2−3=y+52=z−44\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}−3x+2​=2y+5​=4z−4​ is 4430\frac{44}{\sqrt{30}}30​44​, then the largest possible value of ∣λ∣|\lambda|∣λ∣ is equal to ______.

Correct answer: 43

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let P(α,β,γ)P(\alpha, \beta, \gamma)P(α,β,γ) be the image of the point Q(3,−3,1)Q(3, -3, 1)Q(3,−3,1) in the line x−01=y−31=z−1−1\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}1x−0​=1y−3​=−1z−1​ and R be the point (2,5,−1)(2, 5, -1)(2,5,−1). If the area of the triangle PQR is λ\lambdaλ and λ2=14K\lambda^2 = 14Kλ2=14K, then K is equal to:
  1. (A)36
  2. (B)72
  3. (C)18
  4. (D)81

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2024
The shortest distance between the lines x−32=y+15−7=z−95\dfrac{x-3}{2}=\dfrac{y+15}{-7}=\dfrac{z-9}{5}2x−3​=−7y+15​=5z−9​ and x+12=y−11=z−9−3\dfrac{x+1}{2}=\dfrac{y-1}{1}=\dfrac{z-9}{-3}2x+1​=1y−1​=−3z−9​ is
  1. (A)656\sqrt{5}65​
  2. (B)434\sqrt{3}43​
  3. (C)535\sqrt{3}53​
  4. (D)838\sqrt{3}83​

Correct answer: (B)

Step-by-step solution →
Q83·MathematicsNumericalJEE Main 2024
Let PPP be the point (10,−2,−1)(10,-2,-1)(10,−2,−1) and QQQ be the foot of the perpendicular drawn from the point R(1,7,6)R(1,7,6)R(1,7,6) on the line passing through the points (2,−5,11)(2,-5,11)(2,−5,11) and (−6,7,−5)(-6,7,-5)(−6,7,−5). Then the length of the line segment PQPQPQ is equal to _______.

Correct answer: 13

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2024
If the line 2−x3=3y−24λ+1=4−z\dfrac{2-x}{3} = \dfrac{3y-2}{4\lambda+1} = 4 - z32−x​=4λ+13y−2​=4−z makes a right angle with the line x+33μ=1−2y6=5−z7\dfrac{x+3}{3\mu} = \dfrac{1-2y}{6} = \dfrac{5-z}{7}3μx+3​=61−2y​=75−z​, then 4λ+9μ4\lambda + 9\mu4λ+9μ is equal to:
  1. (A)131313
  2. (B)444
  3. (C)555
  4. (D)666

Correct answer: (D)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2024
Let ddd be the distance of the point of intersection of the lines x+63=y2=z+11\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}3x+6​=2y​=1z+1​ and x−74=y−93=z−42\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}4x−7​=3y−9​=2z−4​ from the point (7,8,9)(7, 8, 9)(7,8,9). Then d2+6d^2 + 6d2+6 is equal to:
  1. (A)727272
  2. (B)696969
  3. (C)757575
  4. (D)787878

Correct answer: (C)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2024
Let (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) be the image of the point (8,5,7)(8,5,7)(8,5,7) in the line x−12=y+13=z−25\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{5}2x−1​=3y+1​=5z−2​. Then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)16
  2. (B)18
  3. (C)14
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q87·MathematicsNumericalJEE Main 2024
Let the point (−1,α,β)(-1,\alpha,\beta)(−1,α,β) lie on the line of the shortest distance between the lines x+2−3=y−24=z−52\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}−3x+2​=4y−2​=2z−5​ and x+2−1=y+62=z−10\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}−1x+2​=2y+6​=0z−1​. Then (α−β)2(\alpha-\beta)^2(α−β)2 is equal to __________.

Correct answer: 25

Step-by-step solution →
Q88·MathematicsNumericalJEE Main 2024
Consider a line L passing through the points P(1,2,1)P(1,2,1)P(1,2,1) and Q(2,1,−1)Q(2,1,-1)Q(2,1,−1). If the mirror image of the point A(2,2,2)A(2,2,2)A(2,2,2) in the line L is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then α+β+6γ\alpha+\beta+6\gammaα+β+6γ is equal to

Correct answer: 6

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2024
Let the point, on the line passing through the points P(1,−2,3)P(1,-2,3)P(1,−2,3) and Q(5,−4,7)Q(5,-4,7)Q(5,−4,7), farther from the origin and at a distance of 999 units from the point PPP, be (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). Then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to:
  1. (A)155155155
  2. (B)150150150
  3. (C)160160160
  4. (D)165165165

Correct answer: (A)

Step-by-step solution →
Q90·MathematicsNumericalJEE Main 2024
If the shortest distance between the lines x+22=y+33=z−54\dfrac{x+2}{2}=\dfrac{y+3}{3}=\dfrac{z-5}{4}2x+2​=3y+3​=4z−5​ and x−31=y−2−3=z+42\dfrac{x-3}{1}=\dfrac{y-2}{-3}=\dfrac{z+4}{2}1x−3​=−3y−2​=2z+4​ is 3835k\dfrac{38}{3\sqrt5}k35​38​k and ∫0k[x2] dx=α−α\displaystyle\int_0^k[x^2]\,dx=\alpha-\sqrt\alpha∫0k​[x2]dx=α−α​, where [x][x][x] denotes the greatest integer function, then 6α36\alpha^36α3 is equal to ___

Correct answer: 48

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2024
Let P be the point of intersection of the lines x−21=y−45=z−22\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{2}1x−2​=5y−4​=2z−2​ and x−32=y−23=z−33\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{3}2x−3​=3y−2​=3z−3​. Then the shortest distance of P from the line 4x=2y=z4x=2y=z4x=2y=z is
  1. (A)5147\tfrac{5\sqrt{14}}{7}7514​​
  2. (B)147\tfrac{\sqrt{14}}{7}714​​
  3. (C)3147\tfrac{3\sqrt{14}}{7}7314​​
  4. (D)6147\tfrac{6\sqrt{14}}{7}7614​​

Correct answer: (C)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2024
If the shortest distance between the lines x−λ−2=y−21=z−11\dfrac{x-\lambda}{-2}=\dfrac{y-2}{1}=\dfrac{z-1}{1}−2x−λ​=1y−2​=1z−1​ and x−31=y−1−2=z−22\dfrac{x-\sqrt3}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{2}1x−3​​=−2y−1​=2z−2​ is 111, then the sum of all possible values of λ\lambdaλ is:
  1. (A)000
  2. (B)232\sqrt323​
  3. (C)333\sqrt333​
  4. (D)−23-2\sqrt3−23​

Correct answer: (B)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2024
If the mirror image of the point P(3,4,9)P(3,4,9)P(3,4,9) in the line x−13=y+12=z−21\dfrac{x-1}{3}=\dfrac{y+1}{2}=\dfrac{z-2}{1}3x−1​=2y+1​=1z−2​ is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then 14(α+β+γ)14(\alpha+\beta+\gamma)14(α+β+γ) is:
  1. (A)102
  2. (B)138
  3. (C)108
  4. (D)132

Correct answer: (C)

Step-by-step solution →
Q94·MathematicsNumericalJEE Main 2024
Let the line of the shortest distance between the lines L1:r⃗=(i^+2j^+3k^)+λ(i^−j^+k^)L_1:\vec r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-\hat j+\hat k)L1​:r=(i^+2j^​+3k^)+λ(i^−j^​+k^) and L2:r⃗=(4i^+5j^+6k^)+μ(i^+j^−k^)L_2:\vec r=(4\hat i+5\hat j+6\hat k)+\mu(\hat i+\hat j-\hat k)L2​:r=(4i^+5j^​+6k^)+μ(i^+j^​−k^) intersect L1L_1L1​ and L2L_2L2​ at PPP and QQQ respectively. If (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is the midpoint of the line segment PQPQPQ, then 2(α+β+γ)2(\alpha+\beta+\gamma)2(α+β+γ) is equal to ___

Correct answer: 21

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2024
Let P and Q be the points on the line x+38=y−42=z+12\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}8x+3​=2y−4​=2z+1​ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is:
  1. (A)26
  2. (B)36
  3. (C)18
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2024
The distance of the point Q(0,2,−2)Q(0,2,-2)Q(0,2,−2) from the line passing through the point P(5,−4,3)P(5,-4,3)P(5,−4,3) and perpendicular to the lines r⃗=(−3i^+2k^)+λ(2i^+3j^+5k^)\vec{r}=\left(-3\hat{i}+2\hat{k}\right)+\lambda\left(2\hat{i}+3\hat{j}+5\hat{k}\right)r=(−3i^+2k^)+λ(2i^+3j^​+5k^), λ∈R\lambda\in Rλ∈R and r⃗=(i^−2j^+k^)+μ(−i^+3j^+2k^)\vec{r}=\left(\hat{i}-2\hat{j}+\hat{k}\right)+\mu\left(-\hat{i}+3\hat{j}+2\hat{k}\right)r=(i^−2j^​+k^)+μ(−i^+3j^​+2k^), μ∈R\mu\in Rμ∈R is
  1. (A)86\sqrt{86}86​
  2. (B)20\sqrt{20}20​
  3. (C)54\sqrt{54}54​
  4. (D)74\sqrt{74}74​

Correct answer: (D)

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2024
A line passes through A(4,−6,−2)A(4,-6,-2)A(4,−6,−2) and B(16,−2,4)B(16,-2,4)B(16,−2,4). The point P(a,b,c)P(a,b,c)P(a,b,c) where a,b,ca,b,ca,b,c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a,b,c)P(a,b,c)P(a,b,c) and Q(4,−12,3)Q(4,-12,3)Q(4,−12,3) is equal to ______.

Correct answer: 22

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2024
The shortest distance between lines L1L_1L1​ and L2L_2L2​, where L1:x−12=y+1−3=z+42L_1:\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+4}{2}L1​:2x−1​=−3y+1​=2z+4​ and L2L_2L2​ is the line passing through the points A(−4,4,3)A(-4,4,3)A(−4,4,3), B(−1,6,3)B(-1,6,3)B(−1,6,3) and perpendicular to the line x−3−2=y3=z−11\dfrac{x-3}{-2}=\dfrac{y}{3}=\dfrac{z-1}{1}−2x−3​=3y​=1z−1​, is
  1. (A)121221\dfrac{121}{\sqrt{221}}221​121​
  2. (B)24117\dfrac{24}{\sqrt{117}}117​24​
  3. (C)141221\dfrac{141}{\sqrt{221}}221​141​
  4. (D)42117\dfrac{42}{\sqrt{117}}117​42​

Correct answer: (C)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2024
Let (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) be the mirror image of the point (2,3,5)(2,3,5)(2,3,5) in the line x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}2x−1​=3y−2​=4z−3​. Then 2α+3β+4γ2\alpha+3\beta+4\gamma2α+3β+4γ is equal to
  1. (A)32
  2. (B)33
  3. (C)31
  4. (D)34

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2024
Let Q and R be the feet of perpendiculars from the point P(a,a,a)P(a,a,a)P(a,a,a) on the lines x=yx=yx=y, z=1z=1z=1 and x=−yx=-yx=−y, z=−1z=-1z=−1 respectively. If ∠QPR\angle QPR∠QPR is a right angle, then 12a212a^212a2 is equal to ______.

Correct answer: 12

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2024
Let L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^)L_1:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda(\hat{i}-\hat{j}+2\hat{k})L1​:r=(i^−j^​+2k^)+λ(i^−j^​+2k^), λ∈R\lambda\in Rλ∈R; L2:r⃗=(j^−k^)+μ(3i^+j^+pk^)L_2:\vec{r}=(\hat{j}-\hat{k})+\mu(3\hat{i}+\hat{j}+p\hat{k})L2​:r=(j^​−k^)+μ(3i^+j^​+pk^), μ∈R\mu\in Rμ∈R and L3:r⃗=δ(ℓi^+mj^+nk^)L_3:\vec{r}=\delta(\ell\hat{i}+m\hat{j}+n\hat{k})L3​:r=δ(ℓi^+mj^​+nk^), δ∈R\delta\in Rδ∈R be three lines such that L1L_1L1​ is perpendicular to L2L_2L2​ and L3L_3L3​ is perpendicular to both L1L_1L1​ and L2L_2L2​. Then the point which lies on L3L_3L3​ is:
  1. (A)(−1,7,4)(-1,7,4)(−1,7,4)
  2. (B)(−1,−7,4)(-1,-7,4)(−1,−7,4)
  3. (C)(1,7,−4)(1,7,-4)(1,7,−4)
  4. (D)(1,−7,4)(1,-7,4)(1,−7,4)

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2024
Let (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) be the foot of perpendicular from the point (1,2,3)(1,2,3)(1,2,3) on the line x+35=y−12=z+43\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}5x+3​=2y−1​=3z+4​. Then 19(α+β+γ)19(\alpha+\beta+\gamma)19(α+β+γ) is equal to:
  1. (A)102102102
  2. (B)101101101
  3. (C)999999
  4. (D)100100100

Correct answer: (B)

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2024
If d1d_1d1​ is the shortest distance between the lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z, x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 and d2d_2d2​ is the shortest distance between the lines x−12=y+8−7=z−45\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}2x−1​=−7y+8​=5z−4​, x−12=y−21=z−6−3\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}2x−1​=1y−2​=−3z−6​, then the value of 323 d1d2\dfrac{32\sqrt3\,d_1}{d_2}d2​323​d1​​ is:

Correct answer: 16

Step-by-step solution →
Q104·MathematicsNumericalJEE Main 2024
Let a line passing through the point (−1,2,3)(-1,2,3)(−1,2,3) intersect the lines L1:x−13=y−22=z+1−2L_1:\dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z+1}{-2}L1​:3x−1​=2y−2​=−2z+1​ at M(α,β,γ)M(\alpha,\beta,\gamma)M(α,β,γ) and L2:x+2−3=y−2−2=z−14L_2:\dfrac{x+2}{-3}=\dfrac{y-2}{-2}=\dfrac{z-1}{4}L2​:−3x+2​=−2y−2​=4z−1​ at N(a,b,c)N(a,b,c)N(a,b,c). Then the value of (α+β+γ)2(a+b+c)2\dfrac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}(a+b+c)2(α+β+γ)2​ equals ______.

Correct answer: 196

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2024
Let P(3,2,3)P(3,2,3)P(3,2,3), Q(4,6,2)Q(4,6,2)Q(4,6,2) and R(7,3,2)R(7,3,2)R(7,3,2) be the vertices of △PQR\triangle PQR△PQR. Then, the angle ∠QPR\angle QPR∠QPR is:
  1. (A)π6\dfrac{\pi}{6}6π​
  2. (B)cos⁡−1(718)\cos^{-1}\left(\dfrac{7}{18}\right)cos−1(187​)
  3. (C)cos⁡−1(118)\cos^{-1}\left(\dfrac{1}{18}\right)cos−1(181​)
  4. (D)π3\dfrac{\pi}{3}3π​

Correct answer: (D)

Step-by-step solution →
Q106·MathematicsNumericalJEE Main 2024
Let OOO be the origin, and MMM and NNN be the points on the lines x−54=y−41=z−53\dfrac{x-5}{4}=\dfrac{y-4}{1}=\dfrac{z-5}{3}4x−5​=1y−4​=3z−5​ and x+812=y+25=z+119\dfrac{x+8}{12}=\dfrac{y+2}{5}=\dfrac{z+11}{9}12x+8​=5y+2​=9z+11​ respectively such that MNMNMN is the shortest distance between the given lines. Then OM→⋅ON→\overrightarrow{OM}\cdot\overrightarrow{ON}OM⋅ON is equal to ___.

Correct answer: 9

Step-by-step solution →
Q107·MathematicsNumericalJEE Main 2024
A line with direction ratios 2,1,22,1,22,1,2 meets the lines x=y+2=zx=y+2=zx=y+2=z and x+2=2y=2zx+2=2y=2zx+2=2y=2z respectively at the point P and Q. If the length of the perpendicular from the point (1,2,12)(1,2,12)(1,2,12) to the line PQ is lll, then l2l^2l2 is ______.

Correct answer: 65

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2024
Let PQR be a triangle with R(−1,4,2)R(-1,4,2)R(−1,4,2). Suppose M(2,1,2)M(2,1,2)M(2,1,2) is the mid point of PQ. The distance of the centroid of △PQR\triangle PQR△PQR from the point of intersection of the line x−20=y2=z+3−1\dfrac{x-2}{0}=\dfrac{y}{2}=\dfrac{z+3}{-1}0x−2​=2y​=−1z+3​ and x−11=y+3−3=z+11\dfrac{x-1}{1}=\dfrac{y+3}{-3}=\dfrac{z+1}{1}1x−1​=−3y+3​=1z+1​ is
  1. (A)696969
  2. (B)999
  3. (C)69\sqrt{69}69​
  4. (D)99\sqrt{99}99​

Correct answer: (C)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2024
If the shortest distance between the lines x−41=y+12=z−3\dfrac{x-4}{1}=\dfrac{y+1}{2}=\dfrac{z}{-3}1x−4​=2y+1​=−3z​ and x−λ2=y+14=z−2−5\dfrac{x-\lambda}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{-5}2x−λ​=4y+1​=−5z−2​ is 65\dfrac{6}{\sqrt5}5​6​, then the sum of all possible values of λ\lambdaλ is:
  1. (A)5
  2. (B)8
  3. (C)7
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q110·MathematicsNumericalJEE Main 2024
The lines x−21=y−2=z−78\dfrac{x-2}{1}=\dfrac{y}{-2}=\dfrac{z-7}{8}1x−2​=−2y​=8z−7​ and x+34=y+23=z+21\dfrac{x+3}{4}=\dfrac{y+2}{3}=\dfrac{z+2}{1}4x+3​=3y+2​=1z+2​ intersect at the point P. If the distance of P from the line x+12=y−13=z−11\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z-1}{1}2x+1​=3y−1​=1z−1​ is lll, then 14l214l^214l2 is equal to __________.

Correct answer: 108

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2024
The distance of the point (7,−2,11)(7,-2,11)(7,−2,11) from the line x−61=y−40=z−83\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{3}1x−6​=0y−4​=3z−8​ along the line x−52=y−1−3=z−56\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}2x−5​=−3y−1​=6z−5​, is:
  1. (A)12
  2. (B)14
  3. (C)18
  4. (D)21

Correct answer: (B)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2024
Let the image of the point (1,0,7)(1,0,7)(1,0,7) in the line x1=y−12=z−23\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}1x​=2y−1​=3z−2​ be the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). Then which one of the following points lies on the line passing through (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and making angles 2π3\dfrac{2\pi}{3}32π​ and 3π4\dfrac{3\pi}{4}43π​ with y-axis and z-axis respectively and an acute angle with x-axis ?
  1. (A)(1,−2,1+2)\left(1,-2,1+\sqrt2\right)(1,−2,1+2​)
  2. (B)(1,2,1−2)\left(1,2,1-\sqrt2\right)(1,2,1−2​)
  3. (C)(3,4,3−22)\left(3,4,3-2\sqrt2\right)(3,4,3−22​)
  4. (D)(3,−4,3+22)\left(3,-4,3+2\sqrt2\right)(3,−4,3+22​)

Correct answer: (C)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Advanced 2023
Let Q be the cube with the set of vertices {(x1,x2,x3)∈R3:x1,x2,x3∈{0,1}}\{(x_1, x_2, x_3) \in R^3 : x_1, x_2, x_3 \in \{0, 1\}\}{(x1​,x2​,x3​)∈R3:x1​,x2​,x3​∈{0,1}} . Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0, 0, 0) and (1, 1, 1) is in S. For lines ℓ1\ell_1ℓ1​ and ℓ2\ell_2ℓ2​, let d(ℓ1,ℓ2)d(\ell_1, \ell_2)d(ℓ1​,ℓ2​) denote the shortest distance between them. Then the maximum value of d(ℓ1,ℓ2)d(\ell_1, \ell_2)d(ℓ1​,ℓ2​) as ℓ1\ell_1ℓ1​ varies over F and ℓ2\ell_2ℓ2​ varies over S, is
  1. (A)16\frac{1}{\sqrt{6}}6​1​
  2. (B)18\frac{1}{\sqrt{8}}8​1​
  3. (C)13\frac{1}{\sqrt{3}}3​1​
  4. (D)112\frac{1}{\sqrt{12}}12​1​

Correct answer: (A)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Advanced 2023
Let ℓ1\ell_1ℓ1​ and ℓ2\ell_2ℓ2​ be the lines r1⃗=λ(i^+j^+k^)\vec{r_1} = \lambda\left(\hat{i} + \hat{j} + \hat{k}\right)r1​​=λ(i^+j^​+k^) and r2⃗=(j^−k^)+μ(i^+k^)\vec{r_2} = \left(\hat{j} - \hat{k}\right) + \mu\left(\hat{i} + \hat{k}\right)r2​​=(j^​−k^)+μ(i^+k^), respectively. Let X be the set of all the planes H that contain the line ℓ1\ell_1ℓ1​. For a plane H, let d(H) denote the smallest possible distance between the points of ℓ2\ell_2ℓ2​ and H . Let H0H_0H0​ be a plane in X for which d(H0)d(H_0)d(H0​) is the maximum value of d(H) as H varies over all planes in X . Match each entry in List-I to the correct entries in List-II. The correct option is:
List – IList – II
P.The value of d(H0)d(H_0)d(H0​) is1.3\sqrt{3}3​
Q.The distance of the point (0, 1, 2) from H0H_0H0​ is2.13\frac{1}{\sqrt{3}}3​1​
R.The distance of origin from H0H_0H0​ is3.0
S.The distance of origin from the point of intersection of planes y=zy = zy=z , x=1x = 1x=1 and H0H_0H0​ is4.2\sqrt{2}2​
5.12\frac{1}{\sqrt{2}}2​1​
  1. (A)(P) → (2) (Q) → (4) (R) → (5) (S) → (1)
  2. (B)(P) → (5) (Q) → (4) (R) → (3) (S) → (1)
  3. (C)(P) → (2) (Q) → (1) (R) → (3) (S) → (2)
  4. (D)(P) → (5) (Q) → (1) (R) → (4) (S) → (2)

Correct answer: (B)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2023
Let SSS be the set of all values of λ\lambdaλ, for which the shortest distance between the lines x−λ0=y−34=z+61\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}0x−λ​=4y−3​=1z+6​ and x+λ3=y−4=z−60\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}3x+λ​=−4y​=0z−6​ is 131313. Then 8∣∑λ∈Sλ∣8\left|\sum_{\lambda\in S}\lambda\right|8​∑λ∈S​λ​ is equal to
  1. (A)304
  2. (B)308
  3. (C)306
  4. (D)302

Correct answer: (C)

Step-by-step solution →
Q116·MathematicsNumericalJEE Main 2023
If the line x=y=zx = y = zx=y=z intersects the line xsin⁡A+ysin⁡B+zsin⁡C−18=0=xsin⁡2A+ysin⁡2B+zsin⁡2C−9x\sin A + y\sin B + z\sin C - 18 = 0 = x\sin 2A + y\sin 2B + z\sin 2C - 9xsinA+ysinB+zsinC−18=0=xsin2A+ysin2B+zsin2C−9, where A,B,CA, B, CA,B,C are the angles of a triangle ABCABCABC, then 80(sin⁡A2sin⁡B2sin⁡C2)80\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)80(sin2A​sin2B​sin2C​) is equal to____

Correct answer: 5

Step-by-step solution →
Q117·MathematicsNumericalJEE Main 2023
Let the plane PPP contain the line 2x+y−z−3=0=5x−3y+4z+92x + y - z - 3 = 0 = 5x - 3y + 4z + 92x+y−z−3=0=5x−3y+4z+9 and be parallel to the line x+22=3−y−4=z−75\frac{x + 2}{2} = \frac{3 - y}{-4} = \frac{z - 7}{5}2x+2​=−43−y​=5z−7​. Then the distance of the point A(8,−1,−19)A(8, -1, -19)A(8,−1,−19) from the plane PPP measured parallel to the line x−3=y−54=2−z−12\frac{x}{-3} = \frac{y - 5}{4} = \frac{2 - z}{-12}−3x​=4y−5​=−122−z​ is equal to____

Correct answer: 26

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2023
Let the foot of perpendicular of the point P(3,−2,−9)P(3, -2, -9)P(3,−2,−9) on the plane passing through the points (−1,−2,−3)(-1, -2, -3)(−1,−2,−3), (9,3,4)(9, 3, 4)(9,3,4), (9,−2,1)(9, -2, 1)(9,−2,1) be Q(α,β,γ)Q(\alpha, \beta, \gamma)Q(α,β,γ). Then the distance of QQQ from the origin is:
  1. (A)29\sqrt{29}29​
  2. (B)35\sqrt{35}35​
  3. (C)42\sqrt{42}42​
  4. (D)38\sqrt{38}38​

Correct answer: (C)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2023
The plane, passing through the points (0,−1,2)(0, -1, 2)(0,−1,2) and (−1,2,1)(-1, 2, 1)(−1,2,1) and parallel to the line passing through (5,1,−7)(5, 1, -7)(5,1,−7) and (1,−1,−1)(1, -1, -1)(1,−1,−1), also passes through the point
  1. (A)(1,−2,1)(1, -2, 1)(1,−2,1)
  2. (B)(0,5,−2)(0, 5, -2)(0,5,−2)
  3. (C)(−2,5,0)(-2, 5, 0)(−2,5,0)
  4. (D)(2,0,1)(2, 0, 1)(2,0,1)

Correct answer: (C)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
The line, that is coplanar to the line x+3−3=y−11=z−55\dfrac{x+3}{-3} = \dfrac{y-1}{1} = \dfrac{z-5}{5}−3x+3​=1y−1​=5z−5​, is
  1. (A)x+11=y−22=z−55\dfrac{x+1}{1} = \dfrac{y-2}{2} = \dfrac{z-5}{5}1x+1​=2y−2​=5z−5​
  2. (B)x+1−1=y−22=z−55\dfrac{x+1}{-1} = \dfrac{y-2}{2} = \dfrac{z-5}{5}−1x+1​=2y−2​=5z−5​
  3. (C)x+1−1=y−22=z−54\dfrac{x+1}{-1} = \dfrac{y-2}{2} = \dfrac{z-5}{4}−1x+1​=2y−2​=4z−5​
  4. (D)x−1−1=y−22=z−55\dfrac{x-1}{-1} = \dfrac{y-2}{2} = \dfrac{z-5}{5}−1x−1​=2y−2​=5z−5​

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2023
The distance of the point (−1,2,3)(-1,2,3)(−1,2,3) from the plane r⃗⋅(i^−2j^+3k^)=10\vec r\cdot(\hat i-2\hat j+3\hat k)=10r⋅(i^−2j^​+3k^)=10 parallel to the line of the shortest distance between the lines r⃗=(i^−j^)+λ(2i^+k^)\vec r=(\hat i-\hat j)+\lambda(2\hat i+\hat k)r=(i^−j^​)+λ(2i^+k^) and r⃗=(2i^−j^)+μ(i^−j^+k^)\vec r=(2\hat i-\hat j)+\mu(\hat i-\hat j+\hat k)r=(2i^−j^​)+μ(i^−j^​+k^) is:
  1. (A)363\sqrt636​
  2. (B)353\sqrt535​
  3. (C)262\sqrt626​
  4. (D)252\sqrt525​

Correct answer: (C)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2023
Let the equation of the plane passing through the line of intersection of the planes x+2y+az=2x+2y+az=2x+2y+az=2 and x−y+z=3x-y+z=3x−y+z=3 be 5x−11y+bz=6a−15x-11y+bz=6a-15x−11y+bz=6a−1. For c∈Zc\in\mathbb{Z}c∈Z, if the distance of this plane from the point (a,−c,c)(a,-c,c)(a,−c,c) is 2a\dfrac{2}{\sqrt a}a​2​, then a+bc\dfrac{a+b}{c}ca+b​ is equal to:
  1. (A)−2-2−2
  2. (B)222
  3. (C)−4-4−4
  4. (D)444

Correct answer: (C)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2023
Let N be the foot of perpendicular from the point P(1,−2,3)P(1, -2, 3)P(1,−2,3) on the line passing through the points (4,5,8)(4, 5, 8)(4,5,8) and (1,−7,5)(1, -7, 5)(1,−7,5). Then the distance of N from the plane 2x−2y+z+5=02x - 2y + z + 5 = 02x−2y+z+5=0 is
  1. (A)6
  2. (B)9
  3. (C)7
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2023
Let the image of the point (53,53,83)\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{8}{3}\right)(35​,35​,38​) in the plane x−2y+z−2=0x-2y+z-2=0x−2y+z−2=0 be PPP. If the distance of point Q(6,−2,α), α>0Q(6,-2,\alpha),\,\alpha>0Q(6,−2,α),α>0 from PPP is 131313, then α\alphaα is equal to _____.

Correct answer: 15

Step-by-step solution →
Q125·MathematicsNumericalJEE Main 2023
Let the plane x+3y−2z+6=0x+3y-2z+6=0x+3y−2z+6=0 meet the co-ordinate axes at the points A, B, C. If the orthocentre of the triangle ABC is (α,β,67)\left(\alpha,\beta,\dfrac67\right)(α,β,76​), then 98(α+β)298(\alpha+\beta)^298(α+β)2 is equal to _________.

Correct answer: 288

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2023
Let the plane P: 4x−y+z=104x-y+z=104x−y+z=10 be rotated by an angle π2\dfrac{\pi}{2}2π​ about its line of intersection with the plane x+y−z=4x+y-z=4x+y−z=4. If α\alphaα is the distance of the point (2,3,−4)(2,3,-4)(2,3,−4) from the new position of the plane P, then 35α35\alpha35α is equal to
  1. (A)90
  2. (B)85
  3. (C)126
  4. (D)170

Correct answer: (C)

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2023
Let the lines l1:x+53=y+41=z−α−2l_1:\dfrac{x+5}{3}=\dfrac{y+4}{1}=\dfrac{z-\alpha}{-2}l1​:3x+5​=1y+4​=−2z−α​ and l2:3x+2y+z−2=0=x−3y+2z−13l_2:3x+2y+z-2=0=x-3y+2z-13l2​:3x+2y+z−2=0=x−3y+2z−13 be coplanar. If the point P(a,b,c)(a,b,c)(a,b,c) on l1l_1l1​ is nearest to the point Q(−4,−3,2)(-4,-3,2)(−4,−3,2), then ∣a∣+∣b∣+∣c∣|a|+|b|+|c|∣a∣+∣b∣+∣c∣ is equal to
  1. (A)12
  2. (B)14
  3. (C)10
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2023
Let a⃗\vec{a}a be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^\hat{i} + \hat{j}, \hat{i} + \hat{k}i^+j^​,i^+k^ and i^−j^,j^−k^\hat{i} - \hat{j}, \hat{j} - \hat{k}i^−j^​,j^​−k^. If θ\thetaθ is the angle between the vector a⃗\vec{a}a and the vector b⃗=2i^−2j^+k^\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}b=2i^−2j^​+k^ and a⃗⋅b⃗=6\vec{a} \cdot \vec{b} = 6a⋅b=6, then the ordered pair (θ,∣a⃗×b⃗∣)\left(\theta, |\vec{a} \times \vec{b}|\right)(θ,∣a×b∣) is equal to
  1. (A)(π4,36)\left(\frac{\pi}{4}, 3\sqrt{6}\right)(4π​,36​)
  2. (B)(π3,36)\left(\frac{\pi}{3}, 3\sqrt{6}\right)(3π​,36​)
  3. (C)(π3,6)\left(\frac{\pi}{3}, 6\right)(3π​,6)
  4. (D)(π4,6)\left(\frac{\pi}{4}, 6\right)(4π​,6)

Correct answer: (D)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2023
Let the line ℓ:x=1−y−2=z−3λ\ell:x=\dfrac{1-y}{-2}=\dfrac{z-3}{\lambda}ℓ:x=−21−y​=λz−3​, λ∈R\lambda\in\mathbb{R}λ∈R meet the plane P:x+2y+3z=4P:x+2y+3z=4P:x+2y+3z=4 at the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). If the angle between the line ℓ\ellℓ and the plane PPP is cos⁡−1(514)\cos^{-1}\left(\sqrt{\dfrac{5}{14}}\right)cos−1(145​​), then α+2β+6γ\alpha+2\beta+6\gammaα+2β+6γ is equal to

Correct answer: 11

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2023
Let PPP be the plane passing through the points (5,3,0)(5,3,0)(5,3,0), (13,3,−2)(13,3,-2)(13,3,−2) and (1,6,2)(1,6,2)(1,6,2). For α∈N\alpha\in\mathbb{N}α∈N, if the distances of the points A(3,4,α)A(3,4,\alpha)A(3,4,α) and B(2,α,a)B(2,\alpha,a)B(2,α,a) from the plane PPP are 222 and 333 respectively, then the positive value of aaa is
  1. (A)666
  2. (B)444
  3. (C)333
  4. (D)555

Correct answer: (B)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2023
If equation of the plane that contains the point (−2,3,5)(-2, 3, 5)(−2,3,5) and is perpendicular to each of the planes 2x+4y+5z=82x + 4y + 5z = 82x+4y+5z=8 and 3x−2y+3z=53x - 2y + 3z = 53x−2y+3z=5 is αx+βy+γz+97=0\alpha x + \beta y + \gamma z + 97 = 0αx+βy+γz+97=0 then α+β+γ=\alpha + \beta + \gamma =α+β+γ=
  1. (A)18
  2. (B)17
  3. (C)16
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q132·MathematicsNumericalJEE Main 2023
Let a line lll pass through the origin and be perpendicular to the lines l1:r⃗=(i^−11j^−7k^)+λ(i^+2j^+3k^),λ∈Rl_1 : \vec{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in \mathbb{R}l1​:r=(i^−11j^​−7k^)+λ(i^+2j^​+3k^),λ∈R and l2:r⃗=(−i^+k^)+μ(2i^+2j^+k^),μ∈Rl_2 : \vec{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k}), \mu \in \mathbb{R}l2​:r=(−i^+k^)+μ(2i^+2j^​+k^),μ∈R. If PPP is the point of intersection of lll and l1l_1l1​, and Q(α,β,γ)Q(\alpha, \beta, \gamma)Q(α,β,γ) is the foot of perpendicular from PPP on l2l_2l2​, then 9(α+β+γ)9(\alpha + \beta + \gamma)9(α+β+γ) is equal to _______ .

Correct answer: 5

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2023
Let the line passing through the points P(2,−1,2)P(2,-1,2)P(2,−1,2) and Q(5,3,4)Q(5,3,4)Q(5,3,4) meet the plane x−y+z=4x-y+z=4x−y+z=4 at the point RRR. Then the distance of the point RRR from the plane x+2y+3z+2=0x+2y+3z+2=0x+2y+3z+2=0 measured parallel to the line x−72=y+32=z−21\dfrac{x-7}{2}=\dfrac{y+3}{2}=\dfrac{z-2}{1}2x−7​=2y+3​=1z−2​ is equal to
  1. (A)31\sqrt{31}31​
  2. (B)189\sqrt{189}189​
  3. (C)61\sqrt{61}61​
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2023
Let (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) be the image of the point P(2,3,5)P(2, 3, 5)P(2,3,5) in the plane 2x+y−3z=62x + y - 3z = 62x+y−3z=6. Then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to
  1. (A)101010
  2. (B)555
  3. (C)121212
  4. (D)999

Correct answer: (A)

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2023
Let the image of the point P(1,2,6)P(1,2,6)P(1,2,6) in the plane passing through the points A(1,2,0)A(1,2,0)A(1,2,0), B(1,4,1)B(1,4,1)B(1,4,1) and C(0,5,1)C(0,5,1)C(0,5,1) be Q(α,β,γ)Q(\alpha,\beta,\gamma)Q(α,β,γ). Then (α2+β2+γ2)(\alpha^2+\beta^2+\gamma^2)(α2+β2+γ2) is equal to:
  1. (A)656565
  2. (B)707070
  3. (C)767676
  4. (D)626262

Correct answer: (A)

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2023
Let P be the point of intersection of the line x+33=y+21=1−z2\dfrac{x+3}{3}=\dfrac{y+2}{1}=\dfrac{1-z}{2}3x+3​=1y+2​=21−z​ and the plane x+y+z=2x+y+z=2x+y+z=2. If the distance of the point P from the plane 3x−4y+12z=323x-4y+12z=323x−4y+12z=32 is q, then q and 2q are the roots of the equation
  1. (A)x2−18x−72=0x^2-18x-72=0x2−18x−72=0
  2. (B)x2+18x+72=0x^2+18x+72=0x2+18x+72=0
  3. (C)x2−18x+72=0x^2-18x+72=0x2−18x+72=0
  4. (D)x2+18x−72=0x^2+18x-72=0x2+18x−72=0

Correct answer: (C)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2023
Let the line x1=6−y−2=z+85\frac{x}{1}=\frac{6-y}{-2}=\frac{z+8}{5}1x​=−26−y​=5z+8​ intersect the lines x−54=y−73=z+21\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}4x−5​=3y−7​=1z+2​ and x+36=3−y3=z−61\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}6x+3​=33−y​=1z−6​ at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x−2y+z=142x-2y+z=142x−2y+z=14 is:
  1. (A)444
  2. (B)103\frac{10}{3}310​
  3. (C)333
  4. (D)113\frac{11}{3}311​

Correct answer: (A)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2023
Let two vertices of triangle ABC be (2,4,6)(2,4,6)(2,4,6) and (0,−2,−5)(0,-2,-5)(0,−2,−5), and its centroid be (2,1,−1)(2,1,-1)(2,1,−1). If the image of third vertex in the plane x+2y+4z=11x+2y+4z=11x+2y+4z=11 is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then αβ+βγ+γα\alpha\beta+\beta\gamma+\gamma\alphaαβ+βγ+γα is equal to
  1. (A)727272
  2. (B)747474
  3. (C)767676
  4. (D)707070

Correct answer: (B)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2023
The shortest distance between the lines x+21=y−2=z−52\dfrac{x+2}{1}=\dfrac{y}{-2}=\dfrac{z-5}{2}1x+2​=−2y​=2z−5​ and x−41=y−12=z+30\dfrac{x-4}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{0}1x−4​=2y−1​=0z+3​ is
  1. (A)666
  2. (B)999
  3. (C)777
  4. (D)888

Correct answer: (B)

Step-by-step solution →
Q140·MathematicsNumericalJEE Main 2023
Let the foot of perpendicular from the point A(4,3,1)A(4,3,1)A(4,3,1) on the plane P:x−y+2z+3=0P:x-y+2z+3=0P:x−y+2z+3=0 be N. If B(5,α,β), α,β∈ZB(5,\alpha,\beta),\ \alpha,\beta\in\mathbb{Z}B(5,α,β), α,β∈Z is a point on plane P such that the area of the triangle ABN is 323\sqrt{2}32​, then α2+β2+αβ\alpha^2+\beta^2+\alpha\betaα2+β2+αβ is equal to _______ .

Correct answer: 7

Step-by-step solution →
Q141·MathematicsNumericalJEE Main 2023
Let λ1,λ2\lambda_{1},\lambda_{2}λ1​,λ2​ be the values of λ\lambdaλ for which the points (52,1,λ)\left(\frac{5}{2},1,\lambda\right)(25​,1,λ) and (−2,0,1)(-2,0,1)(−2,0,1) are at equal distance from the plane 2x+3y−6z+7=02x+3y-6z+7=02x+3y−6z+7=0. If λ1>λ2\lambda_{1}>\lambda_{2}λ1​>λ2​, then the distance of the point (λ1−λ2,λ2,λ1)(\lambda_{1}-\lambda_{2},\lambda_{2},\lambda_{1})(λ1​−λ2​,λ2​,λ1​) from the line x−51=y−12=z+72\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}1x−5​=2y−1​=2z+7​ is

Correct answer: 9

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2023
For a,b∈Za,b\in\mathbb{Z}a,b∈Z and ∣a−b∣≤10|a-b|\leq10∣a−b∣≤10, let the angle between the plane P:ax+y−z=bP:ax+y-z=bP:ax+y−z=b and the line l:x−1=a−y=z+1l:x-1=a-y=z+1l:x−1=a−y=z+1 be cos⁡−1(13)\cos^{-1}\left(\dfrac{1}{3}\right)cos−1(31​). If the distance of the point (6,−6,4)(6,-6,4)(6,−6,4) from the plane P is 363\sqrt{6}36​, then a4+b2a^{4}+b^{2}a4+b2 is equal to
  1. (A)25
  2. (B)85
  3. (C)48
  4. (D)32

Correct answer: (D)

Step-by-step solution →
Q143·MathematicsNumericalJEE Main 2023
Let P1P_1P1​ be the plane 3x−y−7z=113x-y-7z=113x−y−7z=11 and P2P_2P2​ be the plane passing through the points (2,−1,0)(2,-1,0)(2,−1,0), (2,0,−1)(2,0,-1)(2,0,−1) and (5,1,1)(5,1,1)(5,1,1). If the foot of the perpendicular drawn from the point (7,4,−1)(7,4,-1)(7,4,−1) on the line of intersection of the planes P1P_1P1​ and P2P_2P2​ is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to

Correct answer: 11

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2023
The shortest distance between the lines x−44=y+25=z+33\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}4x−4​=5y+2​=3z+3​ and x−13=y−34=z−42\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}3x−1​=4y−3​=2z−4​ is
  1. (A)363\sqrt{6}36​
  2. (B)636\sqrt{3}63​
  3. (C)626\sqrt{2}62​
  4. (D)262\sqrt{6}26​

Correct answer: (A)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2023
Let P be the plane passing through the line x−11=y−2−3=z+57\dfrac{x-1}{1}=\dfrac{y-2}{-3}=\dfrac{z+5}{7}1x−1​=−3y−2​=7z+5​ and the point (2,4,−3)(2,4,-3)(2,4,−3). If the image of the point (−1,3,4)(-1,3,4)(−1,3,4) in the plane P is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to
  1. (A)12
  2. (B)11
  3. (C)9
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2023
If the equation of the plane containing the line x+2y+3z−4=0=2x+y−z+5x+2y+3z-4=0=2x+y-z+5x+2y+3z−4=0=2x+y−z+5 and perpendicular to the plane r⃗=(i^−j^)+λ(i^+j^+k^)+μ(i^−2j^+3k^)\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})r=(i^−j^​)+λ(i^+j^​+k^)+μ(i^−2j^​+3k^) is ax+by+cz=4ax+by+cz=4ax+by+cz=4, then (a−b+c)(a-b+c)(a−b+c) is equal to
  1. (A)20
  2. (B)24
  3. (C)22
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2023
One vertex of a rectangular parallelopiped is at the origin OOO and the lengths of its edges along x, yx,\,yx,y and zzz axes are 3, 43,\,43,4 and 555 units respectively. Let PPP be the vertex (3,4,5)(3,4,5)(3,4,5). Then the shortest distance between the diagonal OPOPOP and an edge parallel to the zzz axis, not passing through OOO or PPP, is:
  1. (A)125\dfrac{12}{\sqrt5}5​12​
  2. (B)1255\dfrac{12}{5\sqrt5}55​12​
  3. (C)12512\sqrt5125​
  4. (D)125\dfrac{12}{5}512​

Correct answer: (D)

Step-by-step solution →
Q148·MathematicsNumericalJEE Main 2023
If the lines x−12=2−y−3=z−3α\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}2x−1​=−32−y​=αz−3​ and x−45=y−12=zβ\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}5x−4​=2y−1​=βz​ intersect, then the magnitude of the minimum value of 8αβ8\alpha\beta8αβ is

Correct answer: 18

Step-by-step solution →
Q149·MathematicsNumericalJEE Main 2023
Let QQQ be the foot of perpendicular from the point P(0,2,3)P(0,2,3)P(0,2,3) on the plane 2x−y+z=92x-y+z=92x−y+z=9. If the coordinates of the point RRR are (6,10,7)(6,10,7)(6,10,7), then the square of the area of the triangle PQRPQRPQR is _____.

Correct answer: 594

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2023
A plane PPP contains the line of intersection of the plane r⃗⋅(i^+j^+k^)=6\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k})=6r⋅(i^+j^​+k^)=6 and r⃗⋅(2i^+3j^+4k^)=−5\vec{r}\cdot(2\hat{i}+3\hat{j}+4\hat{k})=-5r⋅(2i^+3j^​+4k^)=−5. If PPP passes through the point (0,2,−2)(0,2,-2)(0,2,−2), then the square of distance of the point (12,12,18)(12,12,18)(12,12,18) from the plane PPP is
  1. (A)620
  2. (B)156
  3. (C)310
  4. (D)144

Correct answer: (A)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2023
Let the line LLL pass through the point (0,1,2)(0,1,2)(0,1,2), intersect the line x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}2x−1​=3y−2​=4z−3​ and be parallel to the plane 2x+y−3z=42x+y-3z=42x+y−3z=4. Then the distance of the point P(1,−9,2)P(1,-9,2)P(1,−9,2) from the line LLL is
  1. (A)9
  2. (B)54\sqrt{54}54​
  3. (C)69\sqrt{69}69​
  4. (D)74\sqrt{74}74​

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2023
If the equation of the plane passing through the line of intersection of the planes 2x−y+z=32x-y+z=32x−y+z=3 and 4x−3y+5z+9=04x-3y+5z+9=04x−3y+5z+9=0 and parallel to the line x+1−2=y+34=z−25\dfrac{x+1}{-2}=\dfrac{y+3}{4}=\dfrac{z-2}{5}−2x+1​=4y+3​=5z−2​ is ax+by+cz+6=0ax+by+cz+6=0ax+by+cz+6=0, then a+b+ca+b+ca+b+c is equal to:
  1. (A)14
  2. (B)12
  3. (C)13
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q153·MathematicsNumericalJEE Main 2023
Let αx+βy+γz=1\alpha x+\beta y+\gamma z=1αx+βy+γz=1 be the equation of a plane through the point (3,−2,5)(3,-2,5)(3,−2,5) and perpendicular to the line joining the points (1,2,3)(1,2,3)(1,2,3) and (−2,3,5)(-2,3,5)(−2,3,5). Then the value of αβγ\alpha\beta\gammaαβγ is equal to

Correct answer: 6

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2023
Let the plane P pass through the intersection of the planes x+3y−z=2x+3y-z=2x+3y−z=2 and x+2y+3z=6x+2y+3z=6x+2y+3z=6 and be perpendicular to the plane 2x+y−z=02x+y-z=02x+y−z=0. If d is the distance of P from the point (−7,1,1)(-7,1,1)(−7,1,1), then d2d^2d2 is equal to:
  1. (A)25083\dfrac{250}{83}83250​
  2. (B)25082\dfrac{250}{82}82250​
  3. (C)1553\dfrac{15}{53}5315​
  4. (D)2583\dfrac{25}{83}8325​

Correct answer: (A)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2023
The shortest distance between the lines x−51=y−22=z−4−3\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}1x−5​=2y−2​=−3z−4​ and x+31=y+54=z−1−5\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}1x+3​=4y+5​=−5z−1​ is
  1. (A)535\sqrt{3}53​
  2. (B)737\sqrt{3}73​
  3. (C)636\sqrt{3}63​
  4. (D)434\sqrt{3}43​

Correct answer: (C)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2023
The point of intersection C of the plane 8x+y+2z=08x+y+2z=08x+y+2z=0 and the line joining the points A(−3,−6,1)A(-3,-6,1)A(−3,−6,1) and B(2,−4,−3)B(2,-4,-3)B(2,−4,−3) divides the line segment AB internally in the ratio k:1. If a, b, c ([a],[b],[c][a],[b],[c][a],[b],[c] are coprime) are the direction ratios of the perpendicular from the point C on the line 1−x1=y+42=z+23\dfrac{1-x}{1}=\dfrac{y+4}{2}=\dfrac{z+2}{3}11−x​=2y+4​=3z+2​, then ∣a+b+c∣|a+b+c|∣a+b+c∣ is equal to

Correct answer: 10

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2023
Let the image of the point P(2,−1,3)P(2,-1,3)P(2,−1,3) in the plane x+2y−z=0x+2y-z=0x+2y−z=0 be QQQ. Then the distance of the plane 3x+2y+z+29=03x+2y+z+29=03x+2y+z+29=0 from the point QQQ is
  1. (A)2427\frac{24\sqrt{2}}{7}7242​​
  2. (B)2142\sqrt{14}214​
  3. (C)3143\sqrt{14}314​
  4. (D)2227\frac{22\sqrt{2}}{7}7222​​

Correct answer: (C)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2023
Let PPP be the plane, passing through the point (1,−1,−5)(1,-1,-5)(1,−1,−5) and perpendicular to the line joining the points (4,1,−3)(4,1,-3)(4,1,−3) and (2,4,3)(2,4,3)(2,4,3). Then the distance of PPP from the point (3,−2,2)(3,-2,2)(3,−2,2) is
  1. (A)555
  2. (B)444
  3. (C)777
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2023
The foot of perpendicular from the origin OOO to a plane PPP which meets the co-ordinate axes at the points A,B,CA,B,CA,B,C is (2,a,4), a∈N(2,a,4),\ a\in\mathbb{N}(2,a,4), a∈N. If the volume of the tetrahedron OABCOABCOABC is 144144144 unit3^33, then which of the following points is NOT on PPP?
  1. (A)(0,6,3)(0,6,3)(0,6,3)
  2. (B)(0,4,4)(0,4,4)(0,4,4)
  3. (C)(2,2,4)(2,2,4)(2,2,4)
  4. (D)(3,0,4)(3,0,4)(3,0,4)

Correct answer: (D)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2023
Let the shortest distance between the lines L:x−5−2=y−λ0=z+λ1, λ≥0L:\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1},\ \lambda\ge 0L:−2x−5​=0y−λ​=1z+λ​, λ≥0 and L1:x+1=y−1=4−zL_1:x+1=y-1=4-zL1​:x+1=y−1=4−z be 262\sqrt{6}26​. If (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) lies on LLL, then which of the following is NOT possible ?
  1. (A)α−2γ=19\alpha-2\gamma=19α−2γ=19
  2. (B)2α+γ=72\alpha+\gamma=72α+γ=7
  3. (C)2α−γ=92\alpha-\gamma=92α−γ=9
  4. (D)α+2γ=24\alpha+2\gamma=24α+2γ=24

Correct answer: (D)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2023
Let the plane P:8x+α1y+α2z+12=0P:8x+\alpha_1 y+\alpha_2 z+12=0P:8x+α1​y+α2​z+12=0 be parallel to the line L:x+22=y−33=z+45L:\dfrac{x+2}{2}=\dfrac{y-3}{3}=\dfrac{z+4}{5}L:2x+2​=3y−3​=5z+4​. If the intercept of PPP on the yyy-axis is 1, then the distance between PPP and LLL is:
  1. (A)72\sqrt{\dfrac72}27​​
  2. (B)27\sqrt{\dfrac27}72​​
  3. (C)614\dfrac{6}{\sqrt{14}}14​6​
  4. (D)14\sqrt{14}14​

Correct answer: (D)

Step-by-step solution →
Q162·MathematicsNumericalJEE Main 2023
Let the line L:x−12=y+1−1=z−31L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}L:2x−1​=−1y+1​=1z−3​ intersect the plane 2x+y+3z=162x + y + 3z = 162x+y+3z=16 at the point PPP. Let the point QQQ be the foot of perpendicular from the point R(1,−1,−3)R(1, -1, -3)R(1,−1,−3) on the line LLL. If α\alphaα is the area of triangle PQRPQRPQR, then α2\alpha^2α2 is equal to

Correct answer: 180

Step-by-step solution →
Q163·MathematicsNumericalJEE Main 2023
Let θ\thetaθ be the angle between the planes P1:r⃗⋅(i^+j^+2k^)=9P_1: \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9P1​:r⋅(i^+j^​+2k^)=9 and P2:r⃗⋅(2i^−j^+k^)=15P_2: \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15P2​:r⋅(2i^−j^​+k^)=15. Let LLL be the line that meets P2P_2P2​ at the point (4,−2,5)(4, -2, 5)(4,−2,5) and makes an angle θ\thetaθ with the normal of P2P_2P2​. If α\alphaα is the angle between LLL and P2P_2P2​, then (tan⁡2θ)(cot⁡2α)(\tan^2 \theta)(\cot^2 \alpha)(tan2θ)(cot2α) is equal to

Correct answer: 9

Step-by-step solution →
Q164·MathematicsNumericalJEE Main 2023
If λ1<λ2\lambda_1<\lambda_2λ1​<λ2​ are two values of λ\lambdaλ such that the angle between the planes P1:r⃗⋅(3i^−5j^+k^)=7P_1:\vec r\cdot(3\hat i-5\hat j+\hat k)=7P1​:r⋅(3i^−5j^​+k^)=7 and P2:r⃗⋅(λi^+j^−3k^)=9P_2:\vec r\cdot(\lambda\hat i+\hat j-3\hat k)=9P2​:r⋅(λi^+j^​−3k^)=9 is sin⁡−1(265)\sin^{-1}\left(\dfrac{2\sqrt6}{5}\right)sin−1(526​​), then the square of the length of perpendicular from the point (38λ1,10λ2,2)(38\lambda_1,10\lambda_2,2)(38λ1​,10λ2​,2) to the plane P1P_1P1​ is

Correct answer: 315

Step-by-step solution →
Q165·MathematicsNumericalJEE Main 2023
If the equation of the plane passing through the point (1,1,2)(1,1,2)(1,1,2) and perpendicular to the line x−3y+2z−1=0=4x−y+zx-3y+2z-1=0=4x-y+zx−3y+2z−1=0=4x−y+z is Ax+By+Cz=1Ax+By+Cz=1Ax+By+Cz=1, then 140(C−B+A)140(C-B+A)140(C−B+A) is equal to

Correct answer: 15

Step-by-step solution →
Q166·MathematicsNumericalJEE Main 2023
Let a line LLL pass through the point P(2,3,1)P(2, 3, 1)P(2,3,1) and be parallel to the line x+3y−2z−2=0=x−y+2zx + 3y - 2z - 2=0=x - y + 2zx+3y−2z−2=0=x−y+2z. If the distance of LLL from the point (5,3,8)(5, 3, 8)(5,3,8) is α\alphaα, then 3α23\alpha^23α2 is equal to _______.

Correct answer: 158

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2023
Let a unit vector OP→\overrightarrow{OP}OP make angles α,β,γ\alpha,\beta,\gammaα,β,γ with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where β∈(0,π2)\beta\in\left(0,\dfrac{\pi}{2}\right)β∈(0,2π​). If OP→\overrightarrow{OP}OP is perpendicular to the plane through points (1,2,3),(2,3,4)(1,2,3),(2,3,4)(1,2,3),(2,3,4) and (1,5,7)(1,5,7)(1,5,7), then which one of the following is true?
  1. (A)α∈(0,π2)\alpha\in\left(0,\dfrac{\pi}{2}\right)α∈(0,2π​) and γ∈(0,π2)\gamma\in\left(0,\dfrac{\pi}{2}\right)γ∈(0,2π​)
  2. (B)α∈(0,π2)\alpha\in\left(0,\dfrac{\pi}{2}\right)α∈(0,2π​) and γ∈(π2,π)\gamma\in\left(\dfrac{\pi}{2},\pi\right)γ∈(2π​,π)
  3. (C)α∈(π2,π)\alpha\in\left(\dfrac{\pi}{2},\pi\right)α∈(2π​,π) and γ∈(π2,π)\gamma\in\left(\dfrac{\pi}{2},\pi\right)γ∈(2π​,π)
  4. (D)α∈(π2,π)\alpha\in\left(\dfrac{\pi}{2},\pi\right)α∈(2π​,π) and γ∈(0,π2)\gamma\in\left(0,\dfrac{\pi}{2}\right)γ∈(0,2π​)

Correct answer: (C)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2023
If a plane passes through the points (−1,k,0),(2,k,−1),(1,1,2)(-1, k, 0), (2, k, -1), (1, 1, 2)(−1,k,0),(2,k,−1),(1,1,2) and is parallel to the line x−11=2y+12=z+1−1\dfrac{x - 1}{1}=\dfrac{2y + 1}{2}=\dfrac{z + 1}{-1}1x−1​=22y+1​=−1z+1​, then the value of k2+1(k−1)(k−2)\dfrac{k^2 + 1}{(k - 1)(k - 2)}(k−1)(k−2)k2+1​ is:
  1. (A)175\dfrac{17}{5}517​
  2. (B)136\dfrac{13}{6}613​
  3. (C)613\dfrac{6}{13}136​
  4. (D)517\dfrac{5}{17}175​

Correct answer: (B)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2023
A vector v⃗\vec{v}v in the first octant is inclined to the x-axis at 60∘60^{\circ}60∘, to the y-axis at 45∘45^{\circ}45∘ and to the z-axis at an acute angle. If a plane passing through the points (2,−1,1)(\sqrt{2}, -1, 1)(2​,−1,1) and (a,b,c)(a, b, c)(a,b,c) is normal to v⃗\vec{v}v, then:
  1. (A)2a+b+c=1\sqrt{2}a + b + c = 12​a+b+c=1
  2. (B)a+2b+c=1a + \sqrt{2}b + c = 1a+2​b+c=1
  3. (C)a+b+2c=1a + b + \sqrt{2}c = 1a+b+2​c=1
  4. (D)2a−b+c=1\sqrt{2}a - b + c = 12​a−b+c=1

Correct answer: (B)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2023
The line l1l_1l1​ passes through the point (2,6,2)(2,6,2)(2,6,2) and is perpendicular to the plane 2x+y−2z=102x+y-2z=102x+y−2z=10. Then the shortest distance between the line l1l_1l1​ and the line x+12=y+4−3=z2\dfrac{x+1}{2}=\dfrac{y+4}{-3}=\dfrac{z}{2}2x+1​=−3y+4​=2z​ is:
  1. (A)113\dfrac{11}{3}311​
  2. (B)193\dfrac{19}{3}319​
  3. (C)777
  4. (D)999

Correct answer: (D)

Step-by-step solution →
Q171·MathematicsNumericalJEE Main 2023
Let the co-ordinates of one vertex of △ABC\triangle ABC△ABC be A(0,2,α)A(0, 2, \alpha)A(0,2,α) and the other two vertices lie on the line x+α5=y−12=z+43\dfrac{x + \alpha}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3}5x+α​=2y−1​=3z+4​. For α∈Z\alpha \in \mathbb{Z}α∈Z, if the area of △ABC\triangle ABC△ABC is 21 sq. units and the line segment BCBCBC has length 2212\sqrt{21}221​ units, then α2\alpha^2α2 is equal to ________ .

Correct answer: 9

Step-by-step solution →
Q172·MathematicsNumericalJEE Main 2023
Let the equation of the plane PPP containing the line x+10=8−y2=zx + 10 = \dfrac{8 - y}{2} = zx+10=28−y​=z be ax+by+3z=2(a+b)ax + by + 3z = 2(a + b)ax+by+3z=2(a+b) and the distance of the plane PPP from the point (1,27,7)(1, 27, 7)(1,27,7) be ccc. Then a2+b2+c2a^2 + b^2 + c^2a2+b2+c2 is equal to ________ .

Correct answer: 355

Step-by-step solution →
Q173·MathematicsSingle correctJEE Main 2023
The shortest distance between the lines x−12=y−21=z−6−3\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}2x−1​=1y−2​=−3z−6​ and x−12=y+8−7=z−45\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}2x−1​=−7y+8​=5z−4​ is:
  1. (A)535\sqrt353​
  2. (B)232\sqrt323​
  3. (C)333\sqrt333​
  4. (D)434\sqrt343​

Correct answer: (D)

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2023
If the lines x−11=y−22=z+31\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{1}1x−1​=2y−2​=1z+3​ and x−a2=y+23=z−31\dfrac{x-a}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{1}2x−a​=3y+2​=1z−3​ intersect at the point PPP, then the distance of the point PPP from the plane z=az=az=a is:
  1. (A)282828
  2. (B)161616
  3. (C)101010
  4. (D)222222

Correct answer: (A)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2023
The plane 2x−y+z=42x-y+z=42x−y+z=4 intersects the line segment joining the points A(a,−2,4)A(a,-2,4)A(a,−2,4) and B(2,b,−3)B(2,b,-3)B(2,b,−3) at the point CCC in the ratio 2:12:12:1 and the distance of the point CCC from the origin is 5\sqrt55​. If ab<0ab<0ab<0 and PPP is the point (a−b, b, 2b−a)(a-b,\,b,\,2b-a)(a−b,b,2b−a) then CP2CP^2CP2 is equal to:
  1. (A)973\dfrac{97}{3}397​
  2. (B)173\dfrac{17}{3}317​
  3. (C)163\dfrac{16}{3}316​
  4. (D)733\dfrac{73}{3}373​

Correct answer: (B)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2023
The foot of perpendicular of the point (2,0,5)(2,0,5)(2,0,5) on the line x+12=y−15=z+1−1\dfrac{x+1}{2} = \dfrac{y-1}{5} = \dfrac{z+1}{-1}2x+1​=5y−1​=−1z+1​ is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ). Then, which of the following is NOT correct?
  1. (A)βγ=−5\dfrac{\beta}{\gamma} = -5γβ​=−5
  2. (B)γα=58\dfrac{\gamma}{\alpha} = \dfrac{5}{8}αγ​=85​
  3. (C)αβ=−8\dfrac{\alpha}{\beta} = -8βα​=−8
  4. (D)αβγ=415\dfrac{\alpha\beta}{\gamma} = \dfrac{4}{15}γαβ​=154​

Correct answer: (A)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2023
Consider the lines L1L_1L1​ and L2L_2L2​ given by L1:x−12=y−31=z−22L_1:\dfrac{x-1}{2}=\dfrac{y-3}{1}=\dfrac{z-2}{2}L1​:2x−1​=1y−3​=2z−2​, L2:x−21=y−22=z−33L_2:\dfrac{x-2}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}L2​:1x−2​=2y−2​=3z−3​. A line L3L_3L3​ having direction ratios 1,−1,−21,-1,-21,−1,−2, intersects L1L_1L1​ and L2L_2L2​ at the points PPP and QQQ respectively. Then the length of line segment PQPQPQ is:
  1. (A)323\sqrt{2}32​
  2. (B)434\sqrt{3}43​
  3. (C)444
  4. (D)262\sqrt{6}26​

Correct answer: (D)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2023
The distance of the point P(4,6,−2)P(4,6,-2)P(4,6,−2) from the line passing through the point (−3,2,3)(-3,2,3)(−3,2,3) and parallel to a line with direction ratios 3,3,−13,3,-13,3,−1 is equal to:
  1. (A)14\sqrt{14}14​
  2. (B)333
  3. (C)6\sqrt{6}6​
  4. (D)232\sqrt{3}23​

Correct answer: (A)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2023
The shortest distance between the lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z and x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 is
  1. (A)32\dfrac{3}{2}23​
  2. (B)222
  3. (C)52\dfrac{5}{2}25​
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2023
If the shortest distance between the line joining the points (1,2,3)(1,2,3)(1,2,3) and (2,3,4)(2,3,4)(2,3,4), and the line x−12=y+1−1=z−20\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-2}{0}2x−1​=−1y+1​=0z−2​ is aaa, then 28a228a^{2}28a2 is equal to

Correct answer: 18

Step-by-step solution →
Q181·MathematicsNumericalJEE Main 2023
Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5x-2y-z-5=0=x+y+3z-5x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z−7=0=2x+3y+z−2x+y+2z-7=0=2x+3y+z-2x+y+2z−7=0=2x+3y+z−2 be ax+by+cz=65ax+by+cz=65ax+by+cz=65. Then the distance of the point (a,b,c)(a,b,c)(a,b,c) from the plane 2x+2y−z+16=02x+2y-z+16=02x+2y−z+16=0 is _______.

Correct answer: 9

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2023
Let the plane containing the line of intersection of the planes P1:x+(λ+4)y+z=1P_1:x+(\lambda+4)y+z=1P1​:x+(λ+4)y+z=1 and P2:2x+y+z=2P_2:2x+y+z=2P2​:2x+y+z=2 pass through the points (0,1,0)(0,1,0)(0,1,0) and (1,0,1)(1,0,1)(1,0,1). Then the distance of the point (2λ,λ,−λ)(2\lambda,\lambda,-\lambda)(2λ,λ,−λ) from the plane P2P_2P2​ is
  1. (A)464\sqrt646​
  2. (B)363\sqrt636​
  3. (C)565\sqrt656​
  4. (D)262\sqrt626​

Correct answer: (B)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2023
If the foot of the perpendicular drawn from (1,9,7)(1,9,7)(1,9,7) to the line passing through the point (3,2,1)(3,2,1)(3,2,1) and parallel to the planes x+2y+z=0x+2y+z=0x+2y+z=0 and 3y−z=33y-z=33y−z=3 is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to
  1. (A)333
  2. (B)111
  3. (C)−1-1−1
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q184·MathematicsNumericalJEE Main 2023
The shortest distance between the lines x−23=y+12=z−62\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-6}{2}3x−2​=2y+1​=2z−6​ and x−63=1−y2=z+80\dfrac{x-6}{3}=\dfrac{1-y}{2}=\dfrac{z+8}{0}3x−6​=21−y​=0z+8​ is equal to _______ .

Correct answer: 14

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2023
The distance of the point (7,−3,−4)(7,-3,-4)(7,−3,−4) from the plane passing through the points (2,−3,1)(2,-3,1)(2,−3,1), (−1,1,−2)(-1,1,-2)(−1,1,−2) and (3,−4,2)(3,-4,2)(3,−4,2) is:
  1. (A)555
  2. (B)444
  3. (C)525\sqrt{2}52​
  4. (D)424\sqrt{2}42​

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsNumericalJEE Main 2023
If the shortest distance between the lines x+62=y−63=z−64\dfrac{x+\sqrt6}{2}=\dfrac{y-\sqrt6}{3}=\dfrac{z-\sqrt6}{4}2x+6​​=3y−6​​=4z−6​​ and x−λ3=y−264=z+265\dfrac{x-\lambda}{3}=\dfrac{y-2\sqrt6}{4}=\dfrac{z+2\sqrt6}{5}3x−λ​=4y−26​​=5z+26​​ is 666, then the square of sum of all possible values of λ\lambdaλ is

Correct answer: 384

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2023
The distance of the point (−1,9,−16)(-1,9,-16)(−1,9,−16) from the plane 2x+3y−z=52x+3y-z=52x+3y−z=5 measured parallel to the line x+43=2−y4=z−312\dfrac{x+4}{3}=\dfrac{2-y}{4}=\dfrac{z-3}{12}3x+4​=42−y​=12z−3​ is
  1. (A)313131
  2. (B)13213\sqrt{2}132​
  3. (C)20220\sqrt{2}202​
  4. (D)262626

Correct answer: (D)

Step-by-step solution →
Q188·MathematicsMultiple correctJEE Advanced 2022
Let SSS be the reflection of a point QQQ with respect to the plane given by r⃗=−(t+p)i^+tj^+(1+p)k^\vec{r} = -\left(t + p\right)\hat{i} + t\hat{j} + \left(1 + p\right)\hat{k}r=−(t+p)i^+tj^​+(1+p)k^ where ttt, ppp are real parameters and i^\hat{i}i^, j^\hat{j}j^​, k^\hat{k}k^ are the unit vectors along the three positive coordinate axes. If the position vectors of QQQ and SSS are 10i^+15j^+20k^10\hat{i} + 15\hat{j} + 20\hat{k}10i^+15j^​+20k^ and αi^+βj^+γk^\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}αi^+βj^​+γk^ respectively, then which of the following is/are TRUE?
  1. (A)3(α+β)=−1013\left(\alpha + \beta\right) = -1013(α+β)=−101
  2. (B)3(β+γ)=−713\left(\beta + \gamma\right) = -713(β+γ)=−71
  3. (C)3(γ+α)=−863\left(\gamma + \alpha\right) = -863(γ+α)=−86
  4. (D)3(α+β+γ)=−1213\left(\alpha + \beta + \gamma\right) = -1213(α+β+γ)=−121

Correct answer: (A), (B), (C)

Step-by-step solution →
Q189·MathematicsMultiple correctJEE Advanced 2022
Let P1P_{1}P1​ and P2P_{2}P2​ be two planes given by P1:10x+15y+12z−60=0P_{1} : 10x + 15y + 12z - 60 = 0P1​:10x+15y+12z−60=0, P2:−2x+5y+4z−20=0P_{2} : -2x + 5y + 4z - 20 = 0P2​:−2x+5y+4z−20=0. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on P1P_{1}P1​ and P2P_{2}P2​?
  1. (A)x−10=y−10=z−15\frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}0x−1​=0y−1​=5z−1​
  2. (B)x−6−5=y2=z3\frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}−5x−6​=2y​=3z​
  3. (C)x−2=y−45=z4\frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}−2x​=5y−4​=4z​
  4. (D)x1=y−4−2=z3\frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}1x​=−2y−4​=3z​

Correct answer: (A), (B), (D)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2022
If the foot of the perpendicular from the point A(−1,4,3)A(-1, 4, 3)A(−1,4,3) on the plane P:2x+my+nz=4P : 2x + my + nz = 4P:2x+my+nz=4, is (−2,72,32)\left(-2, \frac{7}{2}, \frac{3}{2}\right)(−2,27​,23​), then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3,−1,−43, -1, -43,−1,−4, is equal to :
  1. (A)1
  2. (B)26\sqrt{26}26​
  3. (C)222\sqrt{2}22​
  4. (D)14\sqrt{14}14​

Correct answer: (B)

Step-by-step solution →
Q191·MathematicsNumericalJEE Main 2022
Let a line with direction ratios a, –4a, –7 be perpendicular to the lines with direction ratios 3, –1, 2b and b, a, –2. If the point of intersection of the line x+1a2+b2=y−2a2−b2=z1\frac{x+1}{a^{2}+b^{2}} = \frac{y-2}{a^{2}-b^{2}} = \frac{z}{1}a2+b2x+1​=a2−b2y−2​=1z​ and the plane x−y+z=0x - y + z = 0x−y+z=0 is (α, β, γ)(\alpha,\,\beta,\,\gamma)(α,β,γ), then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to __________.

Correct answer: 10

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2022
If (2,3,9)(2, 3, 9)(2,3,9), (5,2,1)(5, 2, 1)(5,2,1), (1,λ,8)(1, \lambda, 8)(1,λ,8) and (λ,2,3)(\lambda, 2, 3)(λ,2,3) are coplanar, then the product of all possible values of λ\lambdaλ is:
  1. (A)212\frac{21}{2}221​
  2. (B)598\frac{59}{8}859​
  3. (C)578\frac{57}{8}857​
  4. (D)958\frac{95}{8}895​

Correct answer: (D)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2022
Let Q be the foot of perpendicular drawn from the point P(1,2,3)P(1, 2, 3)P(1,2,3) to the plane x+2y+z=14x + 2y + z = 14x+2y+z=14. If R is a point on the plane such that ∠PRQ=60∘\angle PRQ = 60^\circ∠PRQ=60∘, then the area of △PQR\triangle PQR△PQR is equal to:
  1. (A)32\frac{\sqrt{3}}{2}23​​
  2. (B)3\sqrt{3}3​
  3. (C)232\sqrt{3}23​
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2022
A plane P is parallel to two lines whose direction ratios are −2,1,−3-2, 1, -3−2,1,−3, and −1,2,−2-1, 2, -2−1,2,−2 and it contains the point (2,2,−2)(2,2,-2)(2,2,−2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α,β,γ\alpha, \beta, \gammaα,β,γ. If V is the volume of the tetrahedron OABC, where O is the origin and p=α+β+γp=\alpha+\beta+\gammap=α+β+γ, then the ordered pair (V,p)(V,p)(V,p) is equal to
  1. (A)(48,−13)(48,-13)(48,−13)
  2. (B)(24,−13)(24,-13)(24,−13)
  3. (C)(48,11)(48,11)(48,11)
  4. (D)(24,−5)(24,-5)(24,−5)

Correct answer: (B)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2022
The foot of the perpendicular from a point on the circle x2+y2=1x^{2} + y^{2} = 1x2+y2=1, z=0z = 0z=0 to the plane 2x+3y+z=62x + 3y + z = 62x+3y+z=6 lies on which one of the following curves ?
  1. (A)(6x+5y−12)2+4(3x+7y−8)2=1(6x + 5y - 12)^{2} + 4(3x + 7y - 8)^{2} = 1(6x+5y−12)2+4(3x+7y−8)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  2. (B)(5x+6y−12)2+4(3x+5y−9)2=1(5x + 6y - 12)^{2} + 4(3x + 5y - 9)^{2} = 1(5x+6y−12)2+4(3x+5y−9)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  3. (C)(6x+5y−14)2+9(3x+5y−7)2=1(6x + 5y - 14)^{2} + 9(3x + 5y - 7)^{2} = 1(6x+5y−14)2+9(3x+5y−7)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  4. (D)(5x+6y−14)2+9(3x+7y−8)2=1(5x + 6y - 14)^{2} + 9(3x + 7y - 8)^{2} = 1(5x+6y−14)2+9(3x+7y−8)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y

Correct answer: (B)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2022
Let the lines x−1λ=y−21=z−32\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}λx−1​=1y−2​=2z−3​ and x+26−2=y+183=z+28λ\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}−2x+26​=3y+18​=λz+28​ be coplanar and P be the plane containing these two lines. Then which of the following points does <b>NOT</b> lies on P?
  1. (A)(0,−2,−2)(0,-2,-2)(0,−2,−2)
  2. (B)(−5,0,−1)(-5,0,-1)(−5,0,−1)
  3. (C)(3,−1,0)(3,-1,0)(3,−1,0)
  4. (D)(0,4,5)(0,4,5)(0,4,5)

Correct answer: (D)

Step-by-step solution →
Q197·MathematicsNumericalJEE Main 2022
Let P(−2, −1, 1) and Q(5617,4317,11117)Q\left( \frac{56}{17}, \frac{43}{17}, \frac{111}{17} \right)Q(1756​,1743​,17111​) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α\alphaα, −1-1−1, β\betaβ, where both α\alphaα and β\betaβ are integers of minimum absolute values, then α2+β2\alpha^{2} + \beta^{2}α2+β2 is equal to __________.

Correct answer: 450

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2022
If the length of the perpendicular drawn from the point P(a,4,2)P(a, 4, 2)P(a,4,2), a>0a > 0a>0 on the line x+12=y−33=z−1−1\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}2x+1​=3y−3​=−1z−1​ is 262\sqrt{6}26​ units and Q(α1,α2,α3)Q(\alpha_{1}, \alpha_{2}, \alpha_{3})Q(α1​,α2​,α3​) is the image of the point P in this line, then a+∑i=13αia + \sum_{i=1}^{3}\alpha_{i}a+∑i=13​αi​ is equal to :
  1. (A)777
  2. (B)888
  3. (C)121212
  4. (D)141414

Correct answer: (B)

Step-by-step solution →
Q199·MathematicsNumericalJEE Main 2022
Let the line x−37=y−2−1=z−3−4\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}7x−3​=−1y−2​=−4z−3​ intersect the plane containing the lines x−41=y+1−2=z1\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}1x−4​=−2y+1​=1z​ and 4ax−y+5z−7a=0=2x−5y−z−34ax-y+5z-7a=0=2x-5y-z-34ax−y+5z−7a=0=2x−5y−z−3, a∈Ra\in\mathbb{R}a∈R at the point P(α,β,γ)P\left(\alpha,\beta,\gamma\right)P(α,β,γ). Then the value of α+β+γ\alpha+\beta+\gammaα+β+γ equals ______.

Correct answer: 12

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2022
If the line of intersection of the planes ax+by=3ax + by = 3ax+by=3 and ax+by+cz=0ax + by + cz = 0ax+by+cz=0, a>0a > 0a>0 makes an angle 30∘30^{\circ}30∘ with the plane y−z+2=0y - z + 2 = 0y−z+2=0, then the direction cosines of the line are :
  1. (A)12,12,0\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 02​1​,2​1​,0
  2. (B)12,−12,0\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 02​1​,2​−1​,0
  3. (C)15,−25,0\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}, 05​1​,−5​2​,0
  4. (D)12,−32,0\frac{1}{2}, -\frac{\sqrt{3}}{2}, 021​,−23​​,0

Correct answer: (B)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2022
If the plane P passes through the intersection of two mutually perpendicular planes 2x + ky − 5z = 1 and 3kx − ky + z = 5, k < 3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is
  1. (A)111\frac{1}{11}111​
  2. (B)511\frac{5}{11}115​
  3. (C)6
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q202·MathematicsNumericalJEE Main 2022
The largest value of a, for which the perpendicular distance of the plane containing the lines r⃗=(i^+j^)+λ(i^+aj^−k^)\vec{r}=\left(\hat{i}+\hat{j}\right)+\lambda\left(\hat{i}+a\hat{j}-\hat{k}\right)r=(i^+j^​)+λ(i^+aj^​−k^) and r⃗=(i^+j^)+μ(−i^+j^−ak^)\vec{r}=\left(\hat{i}+\hat{j}\right)+\mu\left(-\hat{i}+\hat{j}-a\hat{k}\right)r=(i^+j^​)+μ(−i^+j^​−ak^) from the point (2,1,4)(2,1,4)(2,1,4) is 3\sqrt{3}3​, is____________.

Correct answer: 2

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2022
The length of the perpendicular from the point (1, –2, 5) on the line passing through (1, 2, 4) and parallel to the line x + y – z = 0 = x – 2y + 3z – 5 is :
  1. (A)212\sqrt{\dfrac{21}{2}}221​​
  2. (B)92\sqrt{\dfrac{9}{2}}29​​
  3. (C)732\sqrt{\dfrac{73}{2}}273​​
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q204·MathematicsNumericalJEE Main 2022
Let Q and R be two points on the line x+12=y+23=z−12\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}2x+1​=3y+2​=2z−1​ at a distance 26\sqrt{26}26​ from the point P(4, 2, 7). Then the square of the area of the triangle PQR is______________.

Correct answer: 153

Step-by-step solution →
Q205·MathematicsNumericalJEE Main 2022
The plane passing through the line L: ℓx−y+3(1−ℓ)z=1\ell x-y+3(1-\ell)z=1ℓx−y+3(1−ℓ)z=1, x+2y−z=2x+2y-z=2x+2y−z=2 and perpendicular to the plane 3x+2y+z=63x+2y+z=63x+2y+z=6 is 3x−8y+7z=43x-8y+7z=43x−8y+7z=4. If θ\thetaθ is the acute angle between the line L and the y-axis, then 415cos⁡2θ415\cos^{2}\theta415cos2θ is equal to______.

Correct answer: 125

Step-by-step solution →
Q206·MathematicsNumericalJEE Main 2022
The line of shortest distance between the lines x−20=y−11=z1\frac{x-2}{0} = \frac{y-1}{1} = \frac{z}{1}0x−2​=1y−1​=1z​ and x−32=y−52=z−11\frac{x-3}{2} = \frac{y-5}{2} = \frac{z-1}{1}2x−3​=2y−5​=1z−1​ makes an angle of cos⁡−1(227)\cos^{-1}\left( \sqrt{\frac{2}{27}} \right)cos−1(272​​) with the plane P:ax−y−z=0P : ax - y - z = 0P:ax−y−z=0, (a>0)(a > 0)(a>0). If the image of the point (1,1,−5)(1, 1, -5)(1,1,−5) in the plane P is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then α+β−γ\alpha + \beta - \gammaα+β−γ is equal to ________.

Correct answer: 3

Step-by-step solution →
Q207·MathematicsSingle correctJEE Main 2022
Let P be the plane containing the straight line x−39=y+4−1=z−7−5\frac{x - 3}{9} = \frac{y + 4}{-1} = \frac{z - 7}{-5}9x−3​=−1y+4​=−5z−7​ and perpendicular to the plane containing the straight lines x2=y3=z5\frac{x}{2} = \frac{y}{3} = \frac{z}{5}2x​=3y​=5z​ and x3=y7=z8\frac{x}{3} = \frac{y}{7} = \frac{z}{8}3x​=7y​=8z​. If d is the distance of P from the point (2,−5,11)(2, -5, 11)(2,−5,11), then d2d^{2}d2 is equal to :
  1. (A)1472\frac{147}{2}2147​
  2. (B)969696
  3. (C)323\frac{32}{3}332​
  4. (D)545454

Correct answer: (C)

Step-by-step solution →
Q208·MathematicsNumericalJEE Main 2022
Let P1:r⃗⋅(2i^+j^−3k^)=4P_1 : \vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4P1​:r⋅(2i^+j^​−3k^)=4 be a plane. Let P2P_2P2​ be another plane which passes through the points (2, -3, 2) (2, − 2, − 3) and (1, −4, 2). If the direction ratios of the line of intersection of P1P_1P1​ and P2P_2P2​ be 16, α, β, then the value of α + β is equal to _____.

Correct answer: 28

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2022
Let x−23=y+1−2=z+3−1\frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z + 3}{-1}3x−2​=−2y+1​=−1z+3​ lie on the plane px−qy+z=5px - qy + z = 5px−qy+z=5, for some p,q∈Rp, q \in \mathbb{R}p,q∈R. The shortest distance of the plane from the origin is:
  1. (A)3109\sqrt{\frac{3}{109}}1093​​
  2. (B)5142\sqrt{\frac{5}{142}}1425​​
  3. (C)571\sqrt{\frac{5}{71}}715​​
  4. (D)1142\sqrt{\frac{1}{142}}1421​​

Correct answer: (B)

Step-by-step solution →
Q210·MathematicsSingle correctJEE Main 2022
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x+2y+2z=16x + 2y + 2z = 16x+2y+2z=16. Let T be a plane passing through the point Q and contains the line r⃗=−k^+λ(i^+j^+2k^),λ∈R\vec{r} = -\hat{k} + \lambda\left(\hat{i} + \hat{j} + 2\hat{k}\right), \lambda \in \mathbb{R}r=−k^+λ(i^+j^​+2k^),λ∈R. Then, which of the following points lies on T?
  1. (A)(2,1,0)(2, 1, 0)(2,1,0)
  2. (B)(1,2,1)(1, 2, 1)(1,2,1)
  3. (C)(1,2,2)(1, 2, 2)(1,2,2)
  4. (D)(1,3,2)(1, 3, 2)(1,3,2)

Correct answer: (B)

Step-by-step solution →
Q211·MathematicsSingle correctJEE Main 2022
If the mirror image of the point (2, 4, 7) in the plane 3x−y+4z=23x - y + 4z = 23x−y+4z=2 is (a, b, c), the 2a+b+2c2a + b + 2c2a+b+2c is equal to :
  1. (A)54
  2. (B)50
  3. (C)−6
  4. (D)−42

Correct answer: (C)

Step-by-step solution →
Q212·MathematicsSingle correctJEE Main 2022
Let the plane ax+by+cz=dax + by + cz = dax+by+cz=d pass through (2,3,−5)(2, 3, -5)(2,3,−5) and is perpendicular to the planes 2x+y−5z=102x + y - 5z = 102x+y−5z=10 and 3x+5y−7z=123x + 5y - 7z = 123x+5y−7z=12. If a,b,c,da, b, c, da,b,c,d are integers d>0d > 0d>0 and gcd⁡(∣a∣,∣b∣,∣c∣,d)=1\gcd(|a|, |b|, |c|, d) = 1gcd(∣a∣,∣b∣,∣c∣,d)=1, then the value of a+7b+c+20da + 7b + c + 20da+7b+c+20d is equal to
  1. (A)181818
  2. (B)202020
  3. (C)242424
  4. (D)222222

Correct answer: (D)

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2022
If two distinct point Q, R lie on the line of intersection of the planes −x+2y−z=0-x + 2y - z = 0−x+2y−z=0 and 3x−5y+2z=03x - 5y + 2z = 03x−5y+2z=0 and PQ=PR=18\mathrm{PQ}=\mathrm{PR}=\sqrt{18}PQ=PR=18​ where the point P is (1, −2, 3), then the area of the triangle PQR is equal to
  1. (A)2338\frac{2}{3}\sqrt{38}32​38​
  2. (B)4338\frac{4}{3}\sqrt{38}34​38​
  3. (C)8338\frac{8}{3}\sqrt{38}38​38​
  4. (D)1523\sqrt{\frac{152}{3}}3152​​

Correct answer: (B)

Step-by-step solution →
Q214·MathematicsNumericalJEE Main 2022
Let the image of the point P(1,2,3)P(1, 2, 3)P(1,2,3) in the line L:x−63=y−12=z−23L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}L:3x−6​=2y−1​=3z−2​ be QQQ. let R(α,β,γ)R(\alpha, \beta, \gamma)R(α,β,γ) be a point that divides internally the line segment PQPQPQ in the ratio 1:31 : 31:3. Then the value of 22(α+β+γ)22(\alpha + \beta + \gamma)22(α+β+γ) is equal to

Correct answer: 125

Step-by-step solution →
Q215·MathematicsSingle correctJEE Main 2022
Let the plane P:r⃗⋅a⃗=dP : \vec{r} \cdot \vec{a} = dP:r⋅a=d contain the line of intersection of two planes r⃗⋅(i^+3j^−k^)=6\vec{r} \cdot \left(\hat{i} + 3\hat{j} - \hat{k}\right) = 6r⋅(i^+3j^​−k^)=6 and r⃗⋅(−6i^+5j^−k^)=7\vec{r} \cdot \left(-6\hat{i} + 5\hat{j} - \hat{k}\right) = 7r⋅(−6i^+5j^​−k^)=7. If the plane P passes through the point (2, 3, 12)\left(2,\, 3,\, \frac{1}{2}\right)(2,3,21​), then the value of ∣13a⃗∣2d2\frac{|13\vec{a}|^2}{d^2}d2∣13a∣2​ is equal to
  1. (A)90
  2. (B)93
  3. (C)95
  4. (D)97

Correct answer: (B)

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2022
The acute angle between the planes P1P_1P1​ and P2P_2P2​, when P1P_1P1​ and P2P_2P2​ are the planes passing through the intersection of the planes 5x + 8y + 13z − 29 = 0 and 8x − 7y + z − 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is
  1. (A)π3\frac{\pi}{3}3π​
  2. (B)π4\frac{\pi}{4}4π​
  3. (C)π6\frac{\pi}{6}6π​
  4. (D)π12\frac{\pi}{12}12π​

Correct answer: (A)

Step-by-step solution →
Q217·MathematicsSingle correctJEE Main 2022
The shortest distance between the lines x−32=y−23=z−1−1\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-1}2x−3​=3y−2​=−1z−1​ and x+32=y−61=z−53\dfrac{x+3}{2}=\dfrac{y-6}{1}=\dfrac{z-5}{3}2x+3​=1y−6​=3z−5​ is :
  1. (A)185\dfrac{18}{\sqrt{5}}5​18​
  2. (B)2235\dfrac{22}{3\sqrt{5}}35​22​
  3. (C)4635\dfrac{46}{3\sqrt{5}}35​46​
  4. (D)636\sqrt{3}63​

Correct answer: (A)

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2022
If two straight lines whose direction cosines are given by the relations l+m−n=0l + m - n = 0l+m−n=0, 3l2+m2+cnl=03l^{2} + m^{2} + cnl = 03l2+m2+cnl=0 are parallel, then the positive value of c is :
  1. (A)6
  2. (B)4
  3. (C)3
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q219·MathematicsNumericalJEE Main 2022
Let the mirror image of the point (a, b, c) with respect to the plane 3x −-− 4y + 12z + 19 = 0 be (a- 6, β\betaβ, γ\gammaγ). If a + b + c = 5, then 7 β\betaβ - 9 γ\gammaγ is equal to _____________.

Correct answer: 137

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Q220·MathematicsSingle correctJEE Main 2022
Let the foot of the perpendicular from the point (1, 2, 4) on the line x+24=y−12=z+13\dfrac{x+2}{4}=\dfrac{y-1}{2}=\dfrac{z+1}{3}4x+2​=2y−1​=3z+1​ be P. Then the distance of P from the plane 3x+4y+12z+23=03x+4y+12z+23=03x+4y+12z+23=0
  1. (A)555
  2. (B)5013\dfrac{50}{13}1350​
  3. (C)444
  4. (D)6313\dfrac{63}{13}1363​

Correct answer: (A)

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2022
If the two lines l1:x−23=y+1−2l_{1} : \frac{x-2}{3} = \frac{y+1}{-2}l1​:3x−2​=−2y+1​, z=2z = 2z=2 and l2:x−11=2y+3α=z+52l_{2} : \frac{x-1}{1} = \frac{2y+3}{\alpha} = \frac{z+5}{2}l2​:1x−1​=α2y+3​=2z+5​ perpendicular, then an angle between the lines l2l_{2}l2​ and l3:1−x3=2y−1−4=z4l_{3} : \frac{1-x}{3} = \frac{2y-1}{-4} = \frac{z}{4}l3​:31−x​=−42y−1​=4z​ is :
  1. (A)cos⁡−1(294)\cos^{-1}\left(\frac{29}{4}\right)cos−1(429​)
  2. (B)sec⁡−1(294)\sec^{-1}\left(\frac{29}{4}\right)sec−1(429​)
  3. (C)cos⁡−1(229)\cos^{-1}\left(\frac{2}{29}\right)cos−1(292​)
  4. (D)cos⁡−1(229)\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)cos−1(29​2​)

Correct answer: (B)

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Q222·MathematicsSingle correctJEE Main 2022
Let the plane 2x+3y+z+20=02x + 3y + z + 20 = 02x+3y+z+20=0 be rotated through a right angle about its line of intersection with the plane x−3y+5z=8x - 3y + 5z = 8x−3y+5z=8. If the mirror image of the point (2,−12,2)\left(2, -\frac{1}{2}, 2\right)(2,−21​,2) in the rotated plane is B(a, b, c), then :
  1. (A)a8=b5=c−4\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}8a​=5b​=−4c​
  2. (B)a4=b5=c−2\frac{a}{4} = \frac{b}{5} = \frac{c}{-2}4a​=5b​=−2c​
  3. (C)a8=b−5=c4\frac{a}{8} = \frac{b}{-5} = \frac{c}{4}8a​=−5b​=4c​
  4. (D)a4=b5=c2\frac{a}{4} = \frac{b}{5} = \frac{c}{2}4a​=5b​=2c​

Correct answer: (A)

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2022
If the lines r⃗=(i^−j^+k^)+λ(3j^−k^)\vec{r}=\left(\hat{i}-\hat{j}+\hat{k}\right)+\lambda\left(3\hat{j}-\hat{k}\right)r=(i^−j^​+k^)+λ(3j^​−k^) and r⃗=(αi^−j^)+μ(2i^−3k^)\vec{r}=\left(\alpha\hat{i}-\hat{j}\right)+\mu\left(2\hat{i}-3\hat{k}\right)r=(αi^−j^​)+μ(2i^−3k^) are co-planar, then distance of the plane containing these two lines from the point (α, 0, 0) is :
  1. (A)29\frac{2}{9}92​
  2. (B)211\frac{2}{11}112​
  3. (C)411\frac{4}{11}114​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2022
If the plane 2x + y – 5z = 0 is rotated about its line of intersection with the plane 3x – y + 4z – 7 = 0 by an angle of π2\frac{\pi}{2}2π​, then the plane after the rotation passes through the point :
  1. (A)(2, –2, 0)
  2. (B)(–2, 2, 0)
  3. (C)(1, 0, 2)
  4. (D)(–1, 0, –2)

Correct answer: (C)

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Q225·MathematicsSingle correctJEE Main 2022
Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, −1, −1), parallel to the line PQ meets the plane S at R, then QR2QR^2QR2 is equal to:
  1. (A)2
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (B)

Step-by-step solution →
Q226·MathematicsNumericalJEE Main 2022
Let l1l_{1}l1​ be the line in xy-plane with x and y intercepts 18\frac{1}{8}81​ and 142\frac{1}{4\sqrt{2}}42​1​ respectively, and l2l_{2}l2​ be the line in zx-plane with x and z intercepts −18-\frac{1}{8}−81​ and −163-\frac{1}{6\sqrt{3}}−63​1​ respectively. If d is the shortest distance between the line l1l_{1}l1​ and l2l_{2}l2​, then d−2d^{-2}d−2 is equal to

Correct answer: 51

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Q227·MathematicsNumericalJEE Main 2022
Let the lines L1:r⃗=λ(i^+2j^+3k^)L_1 : \vec{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k})L1​:r=λ(i^+2j^​+3k^), λ∈R L2:r⃗=(i^+3j^+k^)+μ(i^+j^+5k^)L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu\left(\hat{i} + \hat{j} + 5\hat{k}\right)L2​:r=(i^+3j^​+k^)+μ(i^+j^​+5k^); μ∈R intersect at the point S. If a plane ax + by − z + d = 0 passes through S and is parallel to both the lines L1L_1L1​ and L2L_2L2​, then the value of a + b + d is equal to ________

Correct answer: 5

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Q228·MathematicsSingle correctJEE Main 2022
Let P be the plane passing through the intersection of the planes r⃗⋅(i^+3j^−k^)=5\vec{r} \cdot \left(\hat{i} + 3\hat{j} - \hat{k}\right) = 5r⋅(i^+3j^​−k^)=5 and r⃗⋅(2i^−j^+k^)=3\vec{r} \cdot \left(2\hat{i} - \hat{j} + \hat{k}\right) = 3r⋅(2i^−j^​+k^)=3, and the point (2,1,−2)(2,1,-2)(2,1,−2). Let the position vectors of the points X and Y be i^−2j^+4k^\hat{i} - 2\hat{j} + 4\hat{k}i^−2j^​+4k^ and 5i^−j^+2k^5\hat{i} - \hat{j} + 2\hat{k}5i^−j^​+2k^ respectively. Then the points
  1. (A)X and X + Y are on the same side of P
  2. (B)Y and Y −-− X are on the opposite sides of P
  3. (C)X and Y are on the opposite sides of P
  4. (D)X + Y and X −-− Y are on the same side of P

Correct answer: (C)

Step-by-step solution →
Q229·MathematicsSingle correctJEE Main 2022
Let the points on the plane P be equidistant from the points (−4,2,1)(-4, 2, 1)(−4,2,1) and (2,−2,3)(2, -2, 3)(2,−2,3). Then the acute angle between the plane P and the plane 2x+y+3z=12x + y + 3z = 12x+y+3z=1 is
  1. (A)π6\dfrac{\pi}{6}6π​
  2. (B)π4\dfrac{\pi}{4}4π​
  3. (C)π3\dfrac{\pi}{3}3π​
  4. (D)5π12\dfrac{5\pi}{12}125π​

Correct answer: (C)

Step-by-step solution →
Q230·MathematicsSingle correctJEE Main 2022
If the shortest distance between the lines x−12=y−23=z−3λ\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{\lambda}2x−1​=3y−2​=λz−3​ and x−21=y−44=z−55\dfrac{x-2}{1}=\dfrac{y-4}{4}=\dfrac{z-5}{5}1x−2​=4y−4​=5z−5​ is 13\dfrac{1}{\sqrt{3}}3​1​, then the sum of all possible values of λ\lambdaλ is :
  1. (A)16
  2. (B)6
  3. (C)12
  4. (D)15

Correct answer: (A)

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Q231·MathematicsNumericalJEE Main 2022
If the shortest distance between the line r⃗=(−i^+3k)+λ(i^−aj^)\vec{r} = \left(-\hat{i} + 3k\right) + \lambda\left(\hat{i} - a\hat{j}\right)r=(−i^+3k)+λ(i^−aj^​) and r⃗=(−j^+2k)+μ(i^−j^+k)\vec{r} = \left(-\hat{j} + 2k\right) + \mu\left(\hat{i} - \hat{j} + k\right)r=(−j^​+2k)+μ(i^−j^​+k) is 23\sqrt{\dfrac{2}{3}}32​​, then the integral value of a is equal to

Correct answer: 2

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Q232·MathematicsNumericalJEE Main 2022
Let a line having direction ratios 1, -4, 2 intersect the lines x−73=y−1−1=z+21\dfrac{x-7}{3}=\dfrac{y-1}{-1}=\dfrac{z+2}{1}3x−7​=−1y−1​=1z+2​ and x2=y−73=z1\dfrac{x}{2}=\dfrac{y-7}{3}=\dfrac{z}{1}2x​=3y−7​=1z​ at the point A and B. Then (AB)2(AB)^{2}(AB)2 is equal to ____ .

Correct answer: 84

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Q233·MathematicsNumericalJEE Advanced 2021
Let α\alphaα, β\betaβ and γ\gammaγ be real numbers such that the system of linear equations x+2y+3z=αx + 2y + 3z = \alphax+2y+3z=α 4x+5y+6z=β4x + 5y + 6z = \beta4x+5y+6z=β 7x+8y+9z=γ−17x + 8y + 9z = \gamma - 17x+8y+9z=γ−1 is consistent. Let ∣M∣|M|∣M∣ represent the determinant of the matrix M=[α2γβ10−101]M = \begin{bmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}M=​αβ−1​210​γ01​​ Let P be the plane containing all those (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P. The value of D is ________.

Correct answer: 1.50

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Q234·MathematicsSingle correctJEE Main 2021
The distance of line 3y − 2z − 1 = 0 = 3x − z + 4 from the point (2, − 1, 6) is :
  1. (A)26\sqrt{26}26​
  2. (B)252\sqrt{5}25​
  3. (C)262\sqrt{6}26​
  4. (D)424\sqrt{2}42​

Correct answer: (C)

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Q235·MathematicsSingle correctJEE Main 2021
Let the acute angle bisector of the two planes x − 2y − 2z + 1 = 0 and 2x − 3y − 6z + 1 = 0 be the plane P. Then which of the following points lies on P?
  1. (A)(3,1,−12)\left(3, 1, -\frac{1}{2}\right)(3,1,−21​)
  2. (B)(−2,0,−12)\left(-2, 0, -\frac{1}{2}\right)(−2,0,−21​)
  3. (C)(0, 2, –4)
  4. (D)(4, 0, – 2)

Correct answer: (B)

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Q236·MathematicsNumericalJEE Main 2021
The square of the distance of the point of intersection of the line x−12=y−23=z+16\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}2x−1​=3y−2​=6z+1​ and the plane 2x−y+z=62x - y + z = 62x−y+z=6 from the point (−1,−1,2)(-1, -1, 2)(−1,−1,2) is ______.

Correct answer: 61

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Q237·MathematicsSingle correctJEE Main 2021
The distance of the point (−1,2,−2)(-1, 2, -2)(−1,2,−2) from the line of intersection of the planes 2x+3y+2z=02x + 3y + 2z = 02x+3y+2z=0 and x−2y+z=0x - 2y + z = 0x−2y+z=0 is :
  1. (A)12\frac{1}{\sqrt{2}}2​1​
  2. (B)52\frac{5}{2}25​
  3. (C)422\frac{\sqrt{42}}{2}242​​
  4. (D)342\frac{\sqrt{34}}{2}234​​

Correct answer: (D)

Step-by-step solution →
Q238·MathematicsSingle correctJEE Main 2021
Let the equation of the plane, that passes through the point (1,4,−3)(1, 4, -3)(1,4,−3) and contains the line of intersection of the planes 3x−2y+4z−7=03x - 2y + 4z - 7 = 03x−2y+4z−7=0 and x+5y−2z+9=0x + 5y - 2z + 9 = 0x+5y−2z+9=0, be αx+βy+γz+3=0\alpha x + \beta y + \gamma z + 3 = 0αx+βy+γz+3=0, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to :
  1. (A)−23-23−23
  2. (B)−15-15−15
  3. (C)23
  4. (D)15

Correct answer: (A)

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Q239·MathematicsNumericalJEE Main 2021
Suppose the line x−2α=y−2−5=z+22\frac{x - 2}{\alpha} = \frac{y - 2}{-5} = \frac{z + 2}{2}αx−2​=−5y−2​=2z+2​ lies on the plane x+3y−2z+β=0x + 3y - 2z + \beta = 0x+3y−2z+β=0. Then (α+β)(\alpha + \beta)(α+β) is equal to ______.

Correct answer: 7

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Q240·MathematicsSingle correctJEE Main 2021
The distance of the point (1,−2,3)(1, -2, 3)(1,−2,3) from the plane x−y+z=5x - y + z = 5x−y+z=5 measured parallel to a line, whose direction ratios are 2,3,−62, 3, -62,3,−6 is :
  1. (A)333
  2. (B)555
  3. (C)222
  4. (D)111

Correct answer: (D)

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Q241·MathematicsSingle correctJEE Main 2021
The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m − n = 0 and mn + nl + lm = 0, is :
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π−cos⁡−1(49)\pi - \cos^{-1}\left(\frac{4}{9}\right)π−cos−1(94​)
  3. (C)cos⁡−1(89)\cos^{-1}\left(\frac{8}{9}\right)cos−1(98​)
  4. (D)π3\frac{\pi}{3}3π​

Correct answer: (A)

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Q242·MathematicsSingle correctJEE Main 2021
Equation of a plane at a distance 221\sqrt{\frac{2}{21}}212​​ from the origin, which contains the line of intersection of the planes x−y−z−1=0x - y - z - 1 = 0x−y−z−1=0 and 2x+y−3z+4=02x + y - 3z + 4 = 02x+y−3z+4=0, is :
  1. (A)3x−y−5z+2=03x - y - 5z + 2 = 03x−y−5z+2=0
  2. (B)3x−4z+3=03x - 4z + 3 = 03x−4z+3=0
  3. (C)−x+2y+2z−3=0-x + 2y + 2z - 3 = 0−x+2y+2z−3=0
  4. (D)4x−y−5z+2=04x - y - 5z + 2 = 04x−y−5z+2=0

Correct answer: (D)

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Q243·MathematicsNumericalJEE Main 2021
Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x − y + z + 3 = 0 and let R (3, 5, γ) be a point of this plane. Then the square of the length of the line segment SR is ___________ .

Correct answer: 72

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Q244·MathematicsSingle correctJEE Main 2021
The equation of the plane passing through the line of intersection of the planes r⃗⋅(i^+j^+k^)=1\vec{r}\cdot\left(\hat{i} + \hat{j} + \hat{k}\right) = 1r⋅(i^+j^​+k^)=1 and r⃗⋅(2i^+3j^−k^)+4=0\vec{r}\cdot\left(2\hat{i} + 3\hat{j} - \hat{k}\right) + 4 = 0r⋅(2i^+3j^​−k^)+4=0 and parallel to the x-axis is:
  1. (A)r⃗⋅(j^−3k^)+6=0\vec{r}\cdot\left(\hat{j} - 3\hat{k}\right) + 6 = 0r⋅(j^​−3k^)+6=0
  2. (B)r⃗⋅(i^+3k^)+6=0\vec{r}\cdot\left(\hat{i} + 3\hat{k}\right) + 6 = 0r⋅(i^+3k^)+6=0
  3. (C)r⃗⋅(i^−3k^)+6=0\vec{r}\cdot\left(\hat{i} - 3\hat{k}\right) + 6 = 0r⋅(i^−3k^)+6=0
  4. (D)r⃗⋅(j^−3k^)−6=0\vec{r}\cdot\left(\hat{j} - 3\hat{k}\right) - 6 = 0r⋅(j^​−3k^)−6=0

Correct answer: (A)

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Q245·MathematicsSingle correctJEE Main 2021
A hall has a square floor of dimension 10m ×\times× 10m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is cos⁡−115\cos^{-1}\frac{1}{5}cos−151​, then the height of the hall (in meters) is :
  1. (A)555
  2. (B)2102\sqrt{10}210​
  3. (C)535\sqrt{3}53​
  4. (D)525\sqrt{2}52​

Correct answer: (D)

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Q246·MathematicsSingle correctJEE Main 2021
A plane P contains the line x+2y+3z+1=0=x−y−z−6x + 2y + 3z + 1 = 0 = x - y - z - 6x+2y+3z+1=0=x−y−z−6, and is perpendicular to the plane −2x+y+z+8=0-2x + y + z + 8 = 0−2x+y+z+8=0. Then which of the following points lies on P ?
  1. (A)(−1,1,2)(-1, 1, 2)(−1,1,2)
  2. (B)(0,1,1)(0, 1, 1)(0,1,1)
  3. (C)(1,0,1)(1, 0, 1)(1,0,1)
  4. (D)(2,−1,1)(2, -1, 1)(2,−1,1)

Correct answer: (B)

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Q247·MathematicsNumericalJEE Main 2021
Let Q be the foot of the perpendicular from the point P(7,−2,13)P(7, -2, 13)P(7,−2,13) on the plane containing the lines x+16=y−17=z−38\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}6x+1​=7y−1​=8z−3​ and x−13=y−25=z−37\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}3x−1​=5y−2​=7z−3​. Then (PQ)2(PQ)^{2}(PQ)2, is equal to ________.

Correct answer: 96

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Q248·MathematicsNumericalJEE Main 2021
Let the line L be the projection of the line x−12=y−31=z−42\frac{x-1}{2} = \frac{y-3}{1} = \frac{z-4}{2}2x−1​=1y−3​=2z−4​ in the plane x−2y−z=3x - 2y - z = 3x−2y−z=3. If d is the distance of the point (0,0,6)(0, 0, 6)(0,0,6) from L, then d2d^{2}d2 is equal to ________.

Correct answer: 26

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Q249·MathematicsSingle correctJEE Main 2021
Let P be the plane passing through the point (1,2,3) and the line of intersection of the planes r⃗⋅(i^+j^+4k^)=16\vec{r} \cdot \left(\hat{i} + \hat{j} + 4\hat{k}\right) = 16r⋅(i^+j^​+4k^)=16 and r⃗⋅(−i^+j^+k^)=6\vec{r} \cdot \left(-\hat{i} + \hat{j} + \hat{k}\right) = 6r⋅(−i^+j^​+k^)=6. Then which of the following points does NOT lie on P ?
  1. (A)(3,3,2)(3, 3, 2)(3,3,2)
  2. (B)(6,−6,2)(6, -6, 2)(6,−6,2)
  3. (C)(4,2,2)(4, 2, 2)(4,2,2)
  4. (D)(−8,8,6)(-8, 8, 6)(−8,8,6)

Correct answer: (C)

Step-by-step solution →
Q250·MathematicsSingle correctJEE Main 2021
For real numbers α and β ≠ 0, if the point of intersection of the straight lines x−α1=y−12=z−13\frac{x - \alpha}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}1x−α​=2y−1​=3z−1​ and x−4β=y−63=z−73\frac{x - 4}{\beta} = \frac{y - 6}{3} = \frac{z - 7}{3}βx−4​=3y−6​=3z−7​, lies on the plane x + 2y − z = 8, then α − β is equal to :
  1. (A)3
  2. (B)9
  3. (C)7
  4. (D)5

Correct answer: (C)

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Q251·MathematicsSingle correctJEE Main 2021
Let the plane passing through the point (−1,0,−2)(-1, 0, -2)(−1,0,−2) and perpendicular to each of the planes 2x+y−z=22x + y - z = 22x+y−z=2 and x−y−z=3x - y - z = 3x−y−z=3 be ax+by+cz+8=0ax + by + cz + 8 = 0ax+by+cz+8=0. Then the value of a+b+ca + b + ca+b+c is equal to :
  1. (A)4
  2. (B)8
  3. (C)5
  4. (D)3

Correct answer: (A)

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Q252·MathematicsNumericalJEE Main 2021
The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,−4,−5) and R(2, −3, 1) and the plane 2x + y + z = 7 , is equal to………….

Correct answer: 7

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Q253·MathematicsNumericalJEE Main 2021
Let a plane P pass through the point (3,7,−7)\left(3,7,-7\right)(3,7,−7) and contain the line, x−2−3=y−32=z+21\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}−3x−2​=2y−3​=1z+2​. If distance of the plane P from the origin is d, then d2d^2d2 is equal to ____.

Correct answer: 3

Step-by-step solution →
Q254·MathematicsNumericalJEE Main 2021
If the lines x−k1=y−22=z−33\dfrac{x-k}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{3}1x−k​=2y−2​=3z−3​ and x+13=y+22=z+31\dfrac{x+1}{3} = \dfrac{y+2}{2} = \dfrac{z+3}{1}3x+1​=2y+2​=1z+3​ are co-planar, then the value of k is......

Correct answer: 1

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Q255·MathematicsSingle correctJEE Main 2021
Let the foot of perpendicular from a point P(1,2,−1)P(1,2,-1)P(1,2,−1) to the straight line L:x1=y0=z−1L:\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}L:1x​=0y​=−1z​ be N. Let a line be drawn from P parallel to the plane x+y+2z=0x+y+2z=0x+y+2z=0 which meets L and point Q. If α\alphaα is the acute angle between the lines PN and PQ, then cos⁡α\cos\alphacosα is equal to……
  1. (A)15\frac{1}{\sqrt{5}}5​1​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)13\frac{1}{\sqrt{3}}3​1​
  4. (D)123\frac{1}{2\sqrt{3}}23​1​

Correct answer: (C)

Step-by-step solution →
Q256·MathematicsSingle correctJEE Main 2021
If the shortest distance between the straight lines 3(x−1)=6(y−2)=2(z−1)3(x - 1) = 6(y - 2) = 2(z - 1)3(x−1)=6(y−2)=2(z−1) and 4(x−2)=2(y−λ)=(z−3)4(x - 2) = 2(y - \lambda) = (z - 3)4(x−2)=2(y−λ)=(z−3), λ∈R\lambda \in Rλ∈R is 138\frac{1}{\sqrt{38}}38​1​, then the integral value of λ\lambdaλ is equal to :
  1. (A)3
  2. (B)2
  3. (C)−1
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q257·MathematicsSingle correctJEE Main 2021
Let L be the line of intersection of planes r⃗.(i^−j^+2k^)=2\vec{r}.\left(\hat{i} - \hat{j} + 2\hat{k}\right) = 2r.(i^−j^​+2k^)=2 and r⃗.(2i^+j^−k^)=2\vec{r}.\left(2\hat{i} + \hat{j} - \hat{k}\right) = 2r.(2i^+j^​−k^)=2. If P(α,β,γ)P(\alpha, \beta, \gamma)P(α,β,γ) is the foot of perpendicular on L from the point (1,2,0)(1, 2, 0)(1,2,0), then the value of 35(α+β+γ)35(\alpha + \beta + \gamma)35(α+β+γ) is equal to :
  1. (A)101101101
  2. (B)143143143
  3. (C)134134134
  4. (D)119119119

Correct answer: (D)

Step-by-step solution →
Q258·MathematicsSingle correctJEE Main 2021
The lines x=ay−1=z−2x=ay-1=z-2x=ay−1=z−2 and x=3y−2=bz−2, (ab≠0)x=3y-2=bz-2,\ (ab\neq 0)x=3y−2=bz−2, (ab=0) are coplanar, if :
  1. (A)a=2, b=2a=2,\ b=2a=2, b=2
  2. (B)b=1, a∈R−{0}b=1,\ a\in R-\{0\}b=1, a∈R−{0}
  3. (C)a=2, b=3a=2,\ b=3a=2, b=3
  4. (D)a=1, b∈R−{0}a=1,\ b\in R-\{0\}a=1, b∈R−{0}

Correct answer: (B)

Step-by-step solution →
Q259·MathematicsSingle correctJEE Main 2021
Consider the line L given by the equation x−32=y−11=z−21\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}2x−3​=1y−1​=1z−2​. Let Q be the mirror image of the point (2,3,−1)(2,3,-1)(2,3,−1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is one the plane P ?
  1. (A)(1,2,2)(1,2,2)(1,2,2)
  2. (B)(−1,1,2)(-1,1,2)(−1,1,2)
  3. (C)(1,1,1)(1,1,1)(1,1,1)
  4. (D)(1,1,2)(1,1,2)(1,1,2)

Correct answer: (A)

Step-by-step solution →
Q260·MathematicsNumericalJEE Main 2021
If the shortest distance between the lines r⃗1=αi^+2j^+2k^+λ(i^−2j^+2k^),λ∈R\vec{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda\left(\hat{i} - 2\hat{j} + 2\hat{k}\right), \lambda \in Rr1​=αi^+2j^​+2k^+λ(i^−2j^​+2k^),λ∈R, α>0\alpha > 0α>0 and r⃗2=−4i^−k^+μ(3i^−2j^−2k^),μ∈R\vec{r}_2 = -4\hat{i} - \hat{k} + \mu\left(3\hat{i} - 2\hat{j} - 2\hat{k}\right), \mu \in Rr2​=−4i^−k^+μ(3i^−2j^​−2k^),μ∈R is 9, then α\alphaα is equal to........

Correct answer: 6

Step-by-step solution →
Q261·MathematicsNumericalJEE Main 2021
Let P be a plane passing through the points (1,0,1)(1,−2,1)\left(1, 0, 1\right)\left(1, -2, 1\right)(1,0,1)(1,−2,1) and (0,1,−2)\left(0, 1, -2\right)(0,1,−2). Let a vector a⃗=αi^+βj^+γk^\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}a=αi^+βj^​+γk^ be such that a⃗\vec{a}a is parallel to the plane P, perpendicular to (i^+2j^+3k^)\left(\hat{i} + 2\hat{j} + 3\hat{k}\right)(i^+2j^​+3k^) and a⃗.(i^+j^+2k^)=2\vec{a}.\left(\hat{i} + \hat{j} + 2\hat{k}\right) = 2a.(i^+j^​+2k^)=2, then (α−β+γ)2\left(\alpha - \beta + \gamma\right)^2(α−β+γ)2 equals........

Correct answer: 81

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Q262·MathematicsNumericalJEE Main 2021
Let P be a plane containing the line x−13=y+64=z+52\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}3x−1​=4y+6​=2z+5​ and parallel to the line x−34=y−2−3=z+57\frac{x-3}{4} = \frac{y-2}{-3} = \frac{z+5}{7}4x−3​=−3y−2​=7z+5​. If the point (1, −1, α) lies on the plane P, then the value of |5α| is equal to ________ .

Correct answer: 38

Step-by-step solution →
Q263·MathematicsNumericalJEE Main 2021
The equation of the planes parallel to the plane x−2y+2z−3=0x - 2y + 2z - 3 = 0x−2y+2z−3=0 which are at unit distance from the point (1,2,3)(1, 2, 3)(1,2,3) is ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0. If (b−d)=K(c−a)(b - d) = K(c - a)(b−d)=K(c−a), then the positive value of K is

Correct answer: 4

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Q264·MathematicsNumericalJEE Main 2021
Let the plane ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0 bisect the line joining the points (4,−3,1)(4, -3, 1)(4,−3,1) and (2,3,−5)(2, 3, -5)(2,3,−5) at the right angles. If a, b, c, d are integers, then the minimum value of (a2+b2+c2+d2)(a^{2} + b^{2} + c^{2} + d^{2})(a2+b2+c2+d2) is

Correct answer: 28

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Q265·MathematicsNumericalJEE Main 2021
Let the mirror image of the point (1, 3, a) with respect to the plane r⃗⋅(2i^−j^+k^)−b=0\vec{r} \cdot \left(2\hat{i} - \hat{j} + \hat{k}\right) - b = 0r⋅(2i^−j^​+k^)−b=0 be (−3, 5, 2). Then the value of |a + b| is equal to ________ .

Correct answer: 1

Step-by-step solution →
Q266·MathematicsNumericalJEE Main 2021
Let P be an arbitrary point having sum of the squares of the distance from the planes x+y+z=0x+y+z=0x+y+z=0, lx−nz=0lx-nz=0lx−nz=0 and x−2y+z=0x-2y+z=0x−2y+z=0, equal to 9. If the locus of the point P is x2+y2+z2=9x^{2}+y^{2}+z^{2}=9x2+y2+z2=9, then the value of l−nl-nl−n is equal to _______.

Correct answer: 0

Step-by-step solution →
Q267·MathematicsNumericalJEE Main 2021
If the equation of the plane passing through the line of intersection of the planes 2x−7y+4z−3=02x - 7y + 4z - 3 = 02x−7y+4z−3=0, 3x−5y+4z+11=03x - 5y + 4z + 11 = 03x−5y+4z+11=0 and the point (−2,1,3)(-2, 1, 3)(−2,1,3) is ax+by+cz−7=0ax + by + cz - 7 = 0ax+by+cz−7=0, then the value of 2a+b+c−72a + b + c - 72a+b+c−7 is ______ .

Correct answer: 4

Step-by-step solution →
Q268·MathematicsSingle correctJEE Main 2021
The equation of the plane which contains the y-axis and passes through the point (1,2,3)(1, 2, 3)(1,2,3) is :
  1. (A)x+3z=10x + 3z = 10x+3z=10
  2. (B)x+3z=0x + 3z = 0x+3z=0
  3. (C)3x+z=63x + z = 63x+z=6
  4. (D)3x−z=03x - z = 03x−z=0

Correct answer: (D)

Step-by-step solution →
Q269·MathematicsSingle correctJEE Main 2021
If the equation of plane passing through the mirror image of a point (2,3,1)(2, 3, 1)(2,3,1) with respect to line x+12=y−31=z+2−1\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}2x+1​=1y−3​=−1z+2​ and containing the line x−23=1−y2=z+11\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}3x−2​=21−y​=1z+1​ is αx+βy+γz=24\alpha x + \beta y + \gamma z = 24αx+βy+γz=24, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to :
  1. (A)202020
  2. (B)191919
  3. (C)181818
  4. (D)212121

Correct answer: (B)

Step-by-step solution →
Q270·MathematicsSingle correctJEE Main 2021
Let P be a plane lx + my + nz = 0 containing the line, 1−x1=y+42=z+23\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}11−x​=2y+4​=3z+2​. If plane P divides the line segment AB joining points A(–3, –6, 1) and B(2, 4, –3) in ratio k : 1 then the value of k is equal to :
  1. (A)1.5
  2. (B)3
  3. (C)2
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q271·MathematicsSingle correctJEE Main 2021
If (x, y, z) be an arbitrary point lying on a plane P which passes through the point (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of expression 3+x−11(y−19)2(z−12)2+y−19(x−11)2(z−12)2+z−12(x−11)2(y−19)2−x+y+z14(x−11)(y−19)(z−12)3+\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\frac{y-19}{(x-11)^{2}(z-12)^{2}}+\frac{z-12}{(x-11)^{2}(y-19)^{2}}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}3+(y−19)2(z−12)2x−11​+(x−11)2(z−12)2y−19​+(x−11)2(y−19)2z−12​−14(x−11)(y−19)(z−12)x+y+z​
  1. (A)000
  2. (B)333
  3. (C)393939
  4. (D)−45-45−45

Correct answer: (B)

Step-by-step solution →
Q272·MathematicsSingle correctJEE Main 2021
If for a > 0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is equal to :
  1. (A)31\sqrt{31}31​
  2. (B)41\sqrt{41}41​
  3. (C)55\sqrt{55}55​
  4. (D)66\sqrt{66}66​

Correct answer: (D)

Step-by-step solution →
Q273·MathematicsSingle correctJEE Main 2021
If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4L_{1}:\frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4}L1​:lx−a​=3y−2​=4z−b​, l≠0l \neq 0l=0 is (3, 5, 7), then the shortest distance between the line L1L_{1}L1​ and line L2:x−23=y−44=z−55L_{2}:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}L2​:3x−2​=4y−4​=5z−5​ is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)16\frac{1}{\sqrt{6}}6​1​
  3. (C)23\sqrt{\frac{2}{3}}32​​
  4. (D)13\frac{1}{\sqrt{3}}3​1​

Correct answer: (B)

Step-by-step solution →
Q274·MathematicsNumericalJEE Main 2021
If the distance of the point (1, -2, 3) from the plane x+2y−3z+10=0x + 2y - 3z + 10 = 0x+2y−3z+10=0 measured parallel to the line, x−13=2−ym=z+31\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}3x−1​=m2−y​=1z+3​ is 72\sqrt{\frac{7}{2}}27​​, then the value of ∣m∣|m|∣m∣ is equal to _______.

Correct answer: 2

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Q275·MathematicsSingle correctJEE Main 2021
If (1,5,35), (7,5,5), (1, λ, 7) and (2λ, 1, 2) are coplanar, then the sum of all possible values of λ is:
  1. (A)−445-\frac{44}{5}−544​
  2. (B)395\frac{39}{5}539​
  3. (C)−395-\frac{39}{5}−539​
  4. (D)445\frac{44}{5}544​

Correct answer: (D)

Step-by-step solution →
Q276·MathematicsSingle correctJEE Main 2021
If the mirror image of the point (1,3,5) with respect to the plane 4x−5y+2z=84x-5y+2z = 84x−5y+2z=8 is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then 5(α+β+γ)5(\alpha + \beta + \gamma)5(α+β+γ) equals:
  1. (A)47
  2. (B)39
  3. (C)43
  4. (D)41

Correct answer: (A)

Step-by-step solution →
Q277·MathematicsSingle correctJEE Main 2021
Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P(α, β, γ) is the foot of perpendicular from (3, 2, 1) on L, then the value of 21(α + β + γ) equals :
  1. (A)142
  2. (B)68
  3. (C)136
  4. (D)102

Correct answer: (D)

Step-by-step solution →
Q278·MathematicsSingle correctJEE Main 2021
Consider the three planes P1P_{1}P1​ : 3x + 15y + 21z = 9, P2P_{2}P2​ : x − 3y − z= 5, and P3P_{3}P3​ : 2x + 10 y + 14z = 5 Then, which one of the following is true?
  1. (A)P1P_{1}P1​ and P3P_{3}P3​ are parallel.
  2. (B)P2P_{2}P2​ and P3P_{3}P3​ are parallel.
  3. (C)P1P_{1}P1​ and P2P_{2}P2​ are parallel.
  4. (D)P1P_{1}P1​ , P2P_{2}P2​ and P3P_{3}P3​ all are parallel.

Correct answer: (A)

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Q279·MathematicsNumericalJEE Main 2021
Let (λ, 2, 1) be a point on the plane which passes through the point (4, −2, 2). If the plane is perpendicular to the line joining the point (−2, −21, 29) and (−1, −16, 23), then (λ11)2−4λ11−4\left(\frac{\lambda}{11}\right)^{2}-\frac{4\lambda}{11}-4(11λ​)2−114λ​−4 is equal to _______.

Correct answer: 8

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Q280·MathematicsNumericalJEE Main 2021
A line ‘λ’ passing through origin is perpendicular to the lines 1_{1}1​ :r=(3+t )ˆi+−+(12t )ˆj+(4+2t)kˆ 2_{2}2​ :r=(3+2s )ˆi+(3+2s )ˆj+(2+s )kˆ If the co-ordinates of the point in the first octant on ‘λ2’ at the distance of 17 from the point of intersection of ‘λ’ and ‘λ1’ are (a, b, c), then 18(a+b+c) is equal to ______.

Correct answer: 44

Step-by-step solution →
Q281·MathematicsSingle correctJEE Main 2021
A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, –1, 1), then the projection of OP on this plane is of length:
  1. (A)2 5
  2. (B)2 3
  3. (C)2 11
  4. (D)2 7

Correct answer: (C)

Step-by-step solution →
Q282·MathematicsSingle correctJEE Main 2021
The equation of the line through the point (0, 1, 2) and perpendicular to the line x−12=y+13=z−1−2\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}2x−1​=3y+1​=−2z−1​ is :
  1. (A)x−3=y−14=z−23\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}−3x​=4y−1​=3z−2​
  2. (B)x3=y−14=z−23\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{3}3x​=4y−1​=3z−2​
  3. (C)x3=y−1−4=z−23\frac{x}{3} = \frac{y-1}{-4} = \frac{z-2}{3}3x​=−4y−1​=3z−2​
  4. (D)x3=y−14=z−2−3\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{-3}3x​=4y−1​=−3z−2​

Correct answer: (A)

Step-by-step solution →
Q283·MathematicsSingle correctJEE Main 2021
Let α\alphaα be the angle between the lines whose direction cosines satisfy the equations l+m−n=0l + m - n = 0l+m−n=0 and l2+m2−n2=0l^{2} + m^{2} - n^{2} = 0l2+m2−n2=0. Then the value of sin⁡4α+cos⁡4α\sin^{4}\alpha + \cos^{4}\alphasin4α+cos4α is :
  1. (A)34\frac{3}{4}43​
  2. (B)12\frac{1}{2}21​
  3. (C)58\frac{5}{8}85​
  4. (D)38\frac{3}{8}83​

Correct answer: (C)

Step-by-step solution →
Q284·MathematicsNumericalJEE Main 2021
Let λ\lambdaλ be an integer. If the shortest distance between the lines x−λ=2y−1=−2zx-\lambda=2y-1=-2zx−λ=2y−1=−2z and x=y+2λ=z−λx=y+2\lambda=z-\lambdax=y+2λ=z−λ is 722\frac{\sqrt{7}}{2\sqrt{2}}22​7​​, then the value of ∣λ∣|\lambda|∣λ∣ is

Correct answer: 1

Step-by-step solution →
Q285·MathematicsSingle correctJEE Main 2021
The equation of the plane passing through the point (1,2,−3)(1, 2, -3)(1,2,−3) and perpendicular to the planes 3x+y−2z=53x + y - 2z = 53x+y−2z=5 and 2x−5y−z=72x - 5y - z = 72x−5y−z=7, is:
  1. (A)3x−10y−2z+11=03x - 10y - 2z + 11 = 03x−10y−2z+11=0
  2. (B)6x−5y−2z−2=06x - 5y - 2z - 2 = 06x−5y−2z−2=0
  3. (C)11x+y+17z+38=011x + y + 17z + 38 = 011x+y+17z+38=0
  4. (D)6x−5y+2z+10=06x - 5y + 2z + 10 = 06x−5y+2z+10=0

Correct answer: (C)

Step-by-step solution →
Q286·MathematicsSingle correctJEE Main 2021
The distance of the point (1,1,9)(1, 1, 9)(1,1,9) from the point of intersection of the line x−31=y−42=z−52\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}1x−3​=2y−4​=2z−5​ and the plane x+y+z=17x + y + z = 17x+y+z=17 is:
  1. (A)38\sqrt{38}38​
  2. (B)19219\sqrt{2}192​
  3. (C)2192\sqrt{19}219​
  4. (D)383838

Correct answer: (A)

Step-by-step solution →
Q287·MathematicsSingle correctJEE Main 2021
The vector equation of the plane passing through the intersection of the planes r⃗⋅(i^+j^+k^)=1\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k})=1r⋅(i^+j^​+k^)=1 and r⃗⋅(i^−2j^)=−2,\vec{r}\cdot(\hat{i}-2\hat{j})=-2,r⋅(i^−2j^​)=−2, and the point (1,0,2) is :
  1. (A)r⃗⋅(i^−7j^+3k^)=73\vec{r}\cdot\left(\hat{i}-7\hat{j}+3\hat{k}\right)=\frac{7}{3}r⋅(i^−7j^​+3k^)=37​
  2. (B)r⃗⋅(i^+7j^+3k^)=7\vec{r}\cdot\left(\hat{i}+7\hat{j}+3\hat{k}\right)=7r⋅(i^+7j^​+3k^)=7
  3. (C)r⃗⋅(3i^+7j^+3k^)=7\vec{r}\cdot\left(3\hat{i}+7\hat{j}+3\hat{k}\right)=7r⋅(3i^+7j^​+3k^)=7
  4. (D)r⃗⋅(i^+7j^+3k^)=73\vec{r}\cdot\left(\hat{i}+7\hat{j}+3\hat{k}\right)=\frac{7}{3}r⋅(i^+7j^​+3k^)=37​

Correct answer: (B)

Step-by-step solution →
Q288·MathematicsSingle correctJEE Main 2021
Let a,b∈Ra,b\in Ra,b∈R. If the mirror image of the point P(a,6,9)P(a,6,9)P(a,6,9) with respect to the line x−37=y−25=z−1−9\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}7x−3​=5y−2​=−9z−1​ is (20,b,−a−9)(20,b,-a-9)(20,b,−a−9), then ∣a+b∣|a+b|∣a+b∣ is equal to :
  1. (A)86
  2. (B)88
  3. (C)84
  4. (D)90

Correct answer: (B)

Step-by-step solution →
Q289·MathematicsMultiple correctJEE Advanced 2020
Let α,β,γ,δ\alpha, \beta, \gamma, \deltaα,β,γ,δ be real numbers such that α2+β2+γ2≠0\alpha^{2} + \beta^{2} + \gamma^{2} \neq 0α2+β2+γ2=0 and α+γ=1\alpha + \gamma = 1α+γ=1. Suppose the point (3,2,−1)(3,2,-1)(3,2,−1) is the mirror image of the point (1,0,−1)(1,0,-1)(1,0,−1) with respect to the plane αx+βy+γz=δ\alpha x + \beta y + \gamma z = \deltaαx+βy+γz=δ. Then which of the following statements is/are TRUE?
  1. (A)α+β=2\alpha + \beta = 2α+β=2
  2. (B)δ−γ=3\delta - \gamma = 3δ−γ=3
  3. (C)δ+β=4\delta + \beta = 4δ+β=4
  4. (D)α+β+γ=δ\alpha + \beta + \gamma = \deltaα+β+γ=δ

Correct answer: (A), (B), (C)

Step-by-step solution →
Q290·MathematicsMultiple correctJEE Advanced 2020
Let L1L_{1}L1​ and L2L_{2}L2​ be the following straight line. L1:x−11=y−1=z−13L_{1} : \frac{x - 1}{1} = \frac{y}{-1} = \frac{z - 1}{3}L1​:1x−1​=−1y​=3z−1​ and L2:x−1−3=y−1=z−11L_{2} : \frac{x - 1}{-3} = \frac{y}{-1} = \frac{z - 1}{1}L2​:−3x−1​=−1y​=1z−1​ Suppose the straight line L:x−αl=y−1m=z−γ−2L : \frac{x - \alpha}{l} = \frac{y - 1}{m} = \frac{z - \gamma}{-2}L:lx−α​=my−1​=−2z−γ​ lies in the plane containing L1L_{1}L1​ and L2L_{2}L2​, and passes through the point of intersection of L1L_{1}L1​ and L2L_{2}L2​. If the line L bisects the acute angle between the lines L1L_{1}L1​ and L2L_{2}L2​, then which of the following statements is/are TRUE?
  1. (A)α−γ=3\alpha - \gamma = 3α−γ=3
  2. (B)l+m=2l + m = 2l+m=2
  3. (C)α−γ=1\alpha - \gamma = 1α−γ=1
  4. (D)l+m=0l + m = 0l+m=0

Correct answer: (A), (B)

Step-by-step solution →
Q291·MathematicsSingle correctJEE Main 2020
A plane P meets the coordinate axes at A, B and C respectively. The centroid of △ABC\triangle ABC△ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is:
  1. (A)x−12=y−11=z−21\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-2}{1}2x−1​=1y−1​=1z−2​
  2. (B)x−11=y−11=z−22\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}1x−1​=1y−1​=2z−2​
  3. (C)x−12=y−12=z−21\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}2x−1​=2y−1​=1z−2​
  4. (D)x−12=y−12=z−22\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{2}2x−1​=2y−1​=2z−2​

Correct answer: (C)

Step-by-step solution →
Q292·MathematicsSingle correctJEE Main 2020
The shortest distance between the lines x−10=y+1−1=z1\dfrac{x-1}{0}=\dfrac{y+1}{-1}=\dfrac{z}{1}0x−1​=−1y+1​=1z​ and x+y+z+1=0,2x−y+z+3=0x+y+z+1=0, 2x-y+z+3=0x+y+z+1=0,2x−y+z+3=0 is:
  1. (A)111
  2. (B)13\dfrac{1}{\sqrt{3}}3​1​
  3. (C)12\dfrac{1}{\sqrt{2}}2​1​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q293·MathematicsSingle correctJEE Main 2020
If (a, b, c) is the image of the point (1, 2, –3) in the line, x+12=y−3−2=z−1\dfrac{x+1}{2} = \dfrac{y-3}{-2} = \dfrac{z}{-1}2x+1​=−2y−3​=−1z​, then a + b+ c is equal to
  1. (A)3
  2. (B)1
  3. (C)2
  4. (D)-1

Correct answer: (C)

Step-by-step solution →
Q294·MathematicsSingle correctJEE Main 2020
If for some α∈R\alpha\in Rα∈R, the lines L1:x+12=y−2−1=z−11L_1:\dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}L1​:2x+1​=−1y−2​=1z−1​ and L2:x+2α=y+15−α=z+11L_2:\dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}L2​:αx+2​=5−αy+1​=1z+1​ are coplanar, then the line L2L_2L2​ passes through the point:
  1. (A)(2,−10,−2)(2,-10,-2)(2,−10,−2)
  2. (B)(10,−2,−2)(10,-2,-2)(10,−2,−2)
  3. (C)(10,2,2)(10,2,2)(10,2,2)
  4. (D)(−2,10,2)(-2,10,2)(−2,10,2)

Correct answer: (A)

Step-by-step solution →
Q295·MathematicsSingle correctJEE Main 2020
The distance of the point (1, –2, 3) from the plane x − y + z = 5 measured parallel to the line x2=y3=z−6\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}2x​=3y​=−6z​ is:
  1. (A)17\frac{1}{7}71​
  2. (B)7
  3. (C)75\frac{7}{5}57​
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q296·MathematicsNumericalJEE Main 2020
If the equation of a plane P, passing through the intersection of the planes, x+4y−z+7=0x + 4y - z + 7 = 0x+4y−z+7=0 and 3x+y+5z=83x + y + 5z = 83x+y+5z=8 is ax+by+6z=15ax + by + 6z = 15ax+by+6z=15 for some a,b∈Ra, b \in Ra,b∈R, then the distance of the point (3,2,−1)(3, 2, -1)(3,2,−1) from the plane P is __________.

Correct answer: 3

Step-by-step solution →
Q297·MathematicsSingle correctJEE Main 2020
The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, −2, 3) and (1, 1, 0) lies on the plane:
  1. (A)x−y−2z=1x-y-2z=1x−y−2z=1
  2. (B)2x+y−z=12x+y-z=12x+y−z=1
  3. (C)x−2y+z=1x-2y+z=1x−2y+z=1
  4. (D)x+2y−z=1x+2y-z=1x+2y−z=1

Correct answer: (B)

Step-by-step solution →
Q298·MathematicsSingle correctJEE Main 2020
The lines r⃗=(i^−j^)+ℓ(2i^+k^)\vec{r}=\left(\hat{i}-\hat{j}\right)+\ell\left(2\hat{i}+\hat{k}\right)r=(i^−j^​)+ℓ(2i^+k^) and r⃗=(2i^−j^)+m(i^+j^−k^)\vec{r}=\left(2\hat{i}-\hat{j}\right)+m\left(\hat{i}+\hat{j}-\hat{k}\right)r=(2i^−j^​)+m(i^+j^​−k^)
  1. (A)intersect when ℓ=2\ell=2ℓ=2 and m=12m=\frac{1}{2}m=21​
  2. (B)intersect when ℓ=1\ell=1ℓ=1 and m=2m=2m=2
  3. (C)do not intersect for any values of ℓ\ellℓ and mmm
  4. (D)intersect for all values of ℓ\ellℓ and mmm

Correct answer: (C)

Step-by-step solution →
Q299·MathematicsSingle correctJEE Main 2020
The plane which bisects the line joining the points (4,−2,3)(4, -2, 3)(4,−2,3) and (2,4,−1)(2, 4, -1)(2,4,−1) at right angles also passes through the point:
  1. (A)(0,−1,1)(0, -1, 1)(0,−1,1)
  2. (B)(4,0,1)(4, 0, 1)(4,0,1)
  3. (C)(4,0,−1)(4, 0, -1)(4,0,−1)
  4. (D)(0,1,−1)(0, 1, -1)(0,1,−1)

Correct answer: (C)

Step-by-step solution →
Q300·MathematicsNumericalJEE Main 2020
Let a plane P contain two lines r⃗=i^+λ(i^+j^),λ∈R\vec{r} = \hat{i} + \lambda \left( \hat{i} + \hat{j} \right), \lambda \in Rr=i^+λ(i^+j^​),λ∈R and r⃗=−j^+μ(j^−k^),μ∈R\vec{r} = -\hat{j} + \mu \left( \hat{j} - \hat{k} \right), \mu \in Rr=−j^​+μ(j^​−k^),μ∈R. If Q(α,β,γ)Q\left( \alpha, \beta, \gamma \right)Q(α,β,γ) is the foot of the perpendicular drawn from the point M (1,0,1)(1, 0, 1)(1,0,1) to P, then 3(α+β+γ)3 \left( \alpha + \beta + \gamma \right)3(α+β+γ) equals ____________ .

Correct answer: 05.00

Step-by-step solution →
Q301·MathematicsNumericalJEE Main 2020
The projection of the line segment joining the points (1,−1,3)(1,-1,3)(1,−1,3) and (2,−4,11)(2,-4,11)(2,−4,11) on the line joining the points (−1,2,3)(-1,2,3)(−1,2,3) and (3,−2,10)(3,-2,10)(3,−2,10) is ______

Correct answer: 8

Step-by-step solution →
Q302·MathematicsSingle correctJEE Main 2020
The mirror image of the point (1, 2, 3) in a plane is (−73,−43,−13)\left(-\dfrac{7}{3},-\dfrac{4}{3},-\dfrac{1}{3}\right)(−37​,−34​,−31​). Which of the following points lies on this plane?
  1. (A)(1, −1, 1)
  2. (B)(1, 1, 1)
  3. (C)(−1, −1, −1)
  4. (D)(−1, −1, 1)

Correct answer: (A)

Step-by-step solution →
Q303·MathematicsSingle correctJEE Main 2020
The shortest distance between the lines x−33=y−8−1=z−31\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}3x−3​=−1y−8​=1z−3​ and x+3−3=y+72=z−64\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}−3x+3​=2y+7​=4z−6​ is:
  1. (A)7230\dfrac{7}{2}\sqrt{30}27​30​
  2. (B)3303\sqrt{30}330​
  3. (C)3
  4. (D)2302\sqrt{30}230​

Correct answer: (B)

Step-by-step solution →
Q304·MathematicsNumericalJEE Main 2020
If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through (α,7,1)(\alpha,7,1)(α,7,1) is (53,73,173)\left(\dfrac{5}{3},\dfrac{7}{3},\dfrac{17}{3}\right)(35​,37​,317​), then α\alphaα is equal to __________.

Correct answer: 4

Step-by-step solution →
Q305·MathematicsSingle correctJEE Main 2020
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is:
  1. (A)(6, 5, 2)
  2. (B)(4, 3, 2)
  3. (C)(6, 5, −2)
  4. (D)(3, 4, −2)

Correct answer: (C)

Step-by-step solution →
Q306·MathematicsMultiple correctJEE Advanced 2019
Let L1L_1L1​ and L2L_2L2​ denote the lines r⃗=i^+λ(−i^+2j^+2k^),λ∈R\vec{r} = \hat{i} + \lambda\left(-\hat{i} + 2\hat{j} + 2\hat{k}\right), \lambda \in Rr=i^+λ(−i^+2j^​+2k^),λ∈R and r⃗=μ(2i^−j^+2k^),μ∈R\vec{r} = \mu\left(2\hat{i} - \hat{j} + 2\hat{k}\right), \mu \in Rr=μ(2i^−j^​+2k^),μ∈R respectively. If L3L_3L3​ is a line which is perpendicular to both L1L_1L1​ and L2L_2L2​ and cuts both of them, then which of the following options describe (s) L3L_3L3​?
  1. (A)r⃗=29(4i^+j^+k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{2}{9}\left(4\hat{i} + \hat{j} + \hat{k}\right) + t\left(2\hat{i} + 2\hat{j} - \hat{k}\right), t \in Rr=92​(4i^+j^​+k^)+t(2i^+2j^​−k^),t∈R
  2. (B)r⃗=13(2i^+k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{1}{3}\left(2\hat{i} + \hat{k}\right) + t\left(2\hat{i} + 2\hat{j} - \hat{k}\right), t \in Rr=31​(2i^+k^)+t(2i^+2j^​−k^),t∈R
  3. (C)r⃗=29(2i^−j^+2k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{2}{9}\left(2\hat{i} - \hat{j} + 2\hat{k}\right) + t\left(2\hat{i} + 2\hat{j} - \hat{k}\right), t \in Rr=92​(2i^−j^​+2k^)+t(2i^+2j^​−k^),t∈R
  4. (D)r⃗=t(2i^+2j^−k^),t∈R\vec{r} = t\left(2\hat{i} + 2\hat{j} - \hat{k}\right), t \in Rr=t(2i^+2j^​−k^),t∈R

Correct answer: (A), (B), (C)

Step-by-step solution →
Q307·MathematicsNumericalJEE Advanced 2019
Three lines are given by r⃗=λi^,λ∈R\vec{r} = \lambda\hat{i}, \lambda \in Rr=λi^,λ∈R, r⃗=μ(i^+j^),μ∈R\vec{r} = \mu\left(\hat{i} + \hat{j}\right), \mu \in Rr=μ(i^+j^​),μ∈R and r⃗=v(i^+j^+k^),v∈R\vec{r} = v\left(\hat{i} + \hat{j} + \hat{k}\right), v \in Rr=v(i^+j^​+k^),v∈R. Let the lines cut the plane x+y+z=1x + y + z = 1x+y+z=1 at the points A, B and C respectively. If the area of the triangle ABC is Δ\DeltaΔ then the value of (6Δ)2(6\Delta)^2(6Δ)2 equals

Correct answer: 0.75

Step-by-step solution →
Q308·MathematicsMultiple correctJEE Advanced 2019
Three lines L1:r⃗=λi^, λ∈RL_1 : \vec{r} = \lambda\hat{i},\ \lambda \in \mathbb{R}L1​:r=λi^, λ∈R L2:r⃗=k^+μj^, μ∈RL_2 : \vec{r} = \hat{k} + \mu\hat{j},\ \mu \in \mathbb{R}L2​:r=k^+μj^​, μ∈R and L3:r⃗=i^+j^+νk^, ν∈RL_3 : \vec{r} = \hat{i} + \hat{j} + \nu\hat{k},\ \nu \in \mathbb{R}L3​:r=i^+j^​+νk^, ν∈R are given. For which point(s) Q on L2L_2L2​ can we find a point P on L1L_1L1​ and a point R on L3L_3L3​ so that P, Q and R are collinear?
  1. (A)k^−12j^\hat{k} - \dfrac{1}{2}\hat{j}k^−21​j^​
  2. (B)k^+12j^\hat{k} + \dfrac{1}{2}\hat{j}k^+21​j^​
  3. (C)k^\hat{k}k^
  4. (D)k^+j^\hat{k} + \hat{j}k^+j^​

Correct answer: (A), (B)

Step-by-step solution →
Q309·MathematicsSingle correctJEE Main 2019
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r⃗=(i^+j^)+λ(i^+2j^−k^)\vec{r} = (\hat{i}+\hat{j}) + \lambda(\hat{i}+2\hat{j}-\hat{k})r=(i^+j^​)+λ(i^+2j^​−k^) and r⃗=(i^+j^)+μ(−i^+j^−2k^)\vec{r} = (\hat{i}+\hat{j}) + \mu(-\hat{i}+\hat{j}-2\hat{k})r=(i^+j^​)+μ(−i^+j^​−2k^) is:
  1. (A)13\dfrac{1}{3}31​
  2. (B)3\sqrt{3}3​
  3. (C)13\dfrac{1}{\sqrt{3}}3​1​
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q310·MathematicsSingle correctJEE Main 2019
A plane which bisects the angle between the two given planes 2x−y+2z−4=02x - y + 2z - 4 = 02x−y+2z−4=0 and x+2y+2z−2=0x + 2y + 2z - 2 = 0x+2y+2z−2=0, passes through the point:
  1. (A)(1, 4, −1)
  2. (B)(2, −4, 1)
  3. (C)(2, 4, 1)
  4. (D)(1, −4, 1)

Correct answer: (B)

Step-by-step solution →
Q311·MathematicsSingle correctJEE Main 2019
The distance of the point having position vector −i^+2j^+6k^-\hat{i} + 2\hat{j} + 6\hat{k}−i^+2j^​+6k^ from the straight line passing through the point (2, 3, -4) and parallel to the vector 6i^+3j^−4k^6\hat{i} + 3\hat{j} - 4\hat{k}6i^+3j^​−4k^ is
  1. (A)7
  2. (B)434\sqrt{3}43​
  3. (C)2132\sqrt{13}213​
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q312·MathematicsSingle correctJEE Main 2019
If the plane 2x−y+2z+3=02x - y + 2z + 3 = 02x−y+2z+3=0 has the distances 13\dfrac{1}{3}31​ and 23\dfrac{2}{3}32​ units from the planes 4x−2y+4z+λ=04x - 2y + 4z + \lambda = 04x−2y+4z+λ=0 and 2x−y+2z+μ=02x - y + 2z + \mu = 02x−y+2z+μ=0, respectively, then the maximum value of λ+μ\lambda + \muλ+μ us equal to
  1. (A)15
  2. (B)13
  3. (C)5
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q313·MathematicsSingle correctJEE Main 2019
A perpendicular is drawn from a point on the line x−12=y+1−1=z1\dfrac{x-1}{2} = \dfrac{y+1}{-1} = \dfrac{z}{1}2x−1​=−1y+1​=1z​ to the plane x+y+z=3x + y + z = 3x+y+z=3 such that the foot of the perpendicular Q also lies on the plane x−y+z=3x - y + z = 3x−y+z=3. Then the co-ordinates of Q are
  1. (A)(2,0,1)(2, 0, 1)(2,0,1)
  2. (B)(−1,0,4)(-1, 0, 4)(−1,0,4)
  3. (C)(1,0,2)(1, 0, 2)(1,0,2)
  4. (D)(4,0,−1)(4, 0, -1)(4,0,−1)

Correct answer: (A)

Step-by-step solution →
Q314·MathematicsSingle correctJEE Main 2019
The vertices B and C of a △ABC\triangle ABC△ABC lie on the line, x+23=y−10=z4\dfrac{x+2}{3}=\dfrac{y-1}{0}=\dfrac{z}{4}3x+2​=0y−1​=4z​ such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A (1,−1,2)(1,-1,2)(1,−1,2) is:
  1. (A)2342\sqrt{34}234​
  2. (B)34\sqrt{34}34​
  3. (C)666
  4. (D)5175\sqrt{17}517​

Correct answer: (B)

Step-by-step solution →
Q315·MathematicsSingle correctJEE Main 2019
A plane passing through the points (0,−1,0)(0,-1,0)(0,−1,0) and (0,0,1)(0,0,1)(0,0,1) and making an angle π4\dfrac{\pi}{4}4π​ with plane y−z+5=0y-z+5=0y−z+5=0, also passes through the point:
  1. (A)(2,1,4)(\sqrt{2},1,4)(2​,1,4)
  2. (B)(−2,−1,−4)(-\sqrt{2},-1,-4)(−2​,−1,−4)
  3. (C)(−2,1,−4)(-\sqrt{2},1,-4)(−2​,1,−4)
  4. (D)(2,−1,4)(\sqrt{2},-1,4)(2​,−1,4)

Correct answer: (A)

Step-by-step solution →
Q316·MathematicsSingle correctJEE Main 2019
If the line, x−12=y+13=z−24\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}2x−1​=3y+1​=4z−2​ meets the plane, x+2y+3z=15x+2y+3z=15x+2y+3z=15 at a point P, then the distance of P from the origin is:
  1. (A)52\dfrac{\sqrt{5}}{2}25​​
  2. (B)252\sqrt{5}25​
  3. (C)92\dfrac{9}{2}29​
  4. (D)72\dfrac{7}{2}27​

Correct answer: (C)

Step-by-step solution →
Q317·MathematicsSingle correctJEE Main 2019
Let P be the plane, which contains the line of intersection of the planes, x+y+z−6=0x+y+z-6=0x+y+z−6=0 and 2x+3y+z+5=02x+3y+z+5=02x+3y+z+5=0 and it is perpendicular to the xy - plane. Then the distance of the point (0,0,256)(0,0,256)(0,0,256) from P is equal to:
  1. (A)63563\sqrt{5}635​
  2. (B)2055205\sqrt{5}2055​
  3. (C)175\dfrac{17}{\sqrt{5}}5​17​
  4. (D)115\dfrac{11}{\sqrt{5}}5​11​

Correct answer: (D)

Step-by-step solution →
Q318·MathematicsSingle correctJEE Main 2019
The length of the perpendicular from the point (2, −1, 4) on the straight line, x+310=y−2−7=z1\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}10x+3​=−7y−2​=1z​ is:
  1. (A)greater than 2 but less than 3
  2. (B)less than 2
  3. (C)greater than 4
  4. (D)greater than 3 but less than 4

Correct answer: (D)

Step-by-step solution →
Q319·MathematicsSingle correctJEE Main 2019
If a point R(4,y,z)R(4,y,z)R(4,y,z) lies on the line segment joining the points P(2,−3,4)P(2,-3,4)P(2,−3,4) and Q(8,0,10)Q(8,0,10)Q(8,0,10), then the distance of R from the origin is:
  1. (A)53\sqrt{53}53​
  2. (B)6
  3. (C)2142\sqrt{14}214​
  4. (D)2212\sqrt{21}221​

Correct answer: (C)

Step-by-step solution →
Q320·MathematicsSingle correctJEE Main 2019
The equation of a plane containing the line of intersection of the planes 2x−y−4=02x-y-4=02x−y−4=0 and y+2z−4=0y+2z-4=0y+2z−4=0 and passing through the point (1, 1, 0) is:
  1. (A)x+3y+z=4x+3y+z=4x+3y+z=4
  2. (B)2x−z=22x-z=22x−z=2
  3. (C)x−3y−2z=−2x-3y-2z=-2x−3y−2z=−2
  4. (D)x−y−z=0x-y-z=0x−y−z=0

Correct answer: (D)

Step-by-step solution →
Q321·MathematicsSingle correctJEE Main 2019
The vector equation of the plane through the line of intersection of the planes x+y+z=1x+y+z=1x+y+z=1 and 2x+3y+4z=52x+3y+4z=52x+3y+4z=5 which is perpendicular to the plane x−y+z=0x-y+z=0x−y+z=0 is:
  1. (A)r⃗×(i^−k^)+2=0\vec{r}\times(\hat{i}-\hat{k})+2=0r×(i^−k^)+2=0
  2. (B)r⃗.(i^−k^)−2=0\vec{r}.(\hat{i}-\hat{k})-2=0r.(i^−k^)−2=0
  3. (C)r⃗.(i^−k^)+2=0\vec{r}.(\hat{i}-\hat{k})+2=0r.(i^−k^)+2=0
  4. (D)x⃗×(i^−k^)+2=0\vec{x}\times(\hat{i}-\hat{k})+2=0x×(i^−k^)+2=0

Correct answer: (C)

Step-by-step solution →
Q322·MathematicsSingle correctJEE Main 2019
A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is:
  1. (A)cos⁡−1(1731)\cos^{-1}\left(\dfrac{17}{31}\right)cos−1(3117​)
  2. (B)cos⁡−1(1935)\cos^{-1}\left(\dfrac{19}{35}\right)cos−1(3519​)
  3. (C)cos⁡−1(935)\cos^{-1}\left(\dfrac{9}{35}\right)cos−1(359​)
  4. (D)cos⁡−1(731)\cos^{-1}\left(\dfrac{7}{31}\right)cos−1(317​)

Correct answer: (B)

Step-by-step solution →
Q323·MathematicsSingle correctJEE Main 2019
Let S be the set of all real values of λ such that a plane passing through the points (−λ2,1,1)(-\lambda^{2}, 1, 1)(−λ2,1,1), (1,−λ2,1)(1, -\lambda^{2}, 1)(1,−λ2,1) and (1,1,−λ2)(1, 1, -\lambda^{2})(1,1,−λ2) also passes through the point (−1, −1, 1). Then S is equal to:
  1. (A){3}\{\sqrt{3}\}{3​}
  2. (B){3,−3}\{\sqrt{3}, -\sqrt{3}\}{3​,−3​}
  3. (C){1,−1}\{1, -1\}{1,−1}
  4. (D){3,−3}\{3, -3\}{3,−3}

Correct answer: (B)

Step-by-step solution →
Q324·MathematicsSingle correctJEE Main 2019
If an angle between the line, x+12=y−21=z−3−2\dfrac{x+1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{-2}2x+1​=1y−2​=−2z−3​ and the plane, x − 2y − kz = 3 is cos⁡−1(223)\cos^{-1}\left(\dfrac{2\sqrt{2}}{3}\right)cos−1(322​​), then a value of k is :
  1. (A)53\sqrt{\dfrac{5}{3}}35​​
  2. (B)35\sqrt{\dfrac{3}{5}}53​​
  3. (C)−35-\dfrac{3}{5}−53​
  4. (D)−53-\dfrac{5}{3}−35​

Correct answer: (A)

Step-by-step solution →
Q325·MathematicsSingle correctJEE Main 2019
The perpendicular distance from the origin to the plane containing the two lines, x+23=y−25=z+57\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}3x+2​=5y−2​=7z+5​ and x−11=y−44=z+47\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}1x−1​=4y−4​=7z+4​, is:
  1. (A)11611\sqrt{6}116​
  2. (B)11/611/\sqrt{6}11/6​
  3. (C)11
  4. (D)6116\sqrt{11}611​

Correct answer: (B)

Step-by-step solution →
Q326·MathematicsSingle correctJEE Main 2019
The plane containing the line x−32=y+2−1=z−13\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}2x−3​=−1y+2​=3z−1​ and also containing its projection on the plane 2x+3y−z=52x+3y-z=52x+3y−z=5, contains which one of the following points?
  1. (A)(2,2,0)(2,2,0)(2,2,0)
  2. (B)(−2,2,2)(-2,2,2)(−2,2,2)
  3. (C)(0,−2,2)(0,-2,2)(0,−2,2)
  4. (D)(2,0,−2)(2,0,-2)(2,0,−2)

Correct answer: (D)

Step-by-step solution →
Q327·MathematicsSingle correctJEE Main 2019
If the point (2, α, β) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x−5y=152x-5y=152x−5y=15, then 2α−3β2\alpha-3\beta2α−3β is equal to:
  1. (A)12
  2. (B)7
  3. (C)5
  4. (D)17

Correct answer: (B)

Step-by-step solution →
Q328·MathematicsSingle correctJEE Main 2019
Two lines x−31=y+13=z−6−1\dfrac{x-3}{1}=\dfrac{y+1}{3}=\dfrac{z-6}{-1}1x−3​=3y+1​=−1z−6​ and x+57=y−2−6=z−34\dfrac{x+5}{7}=\dfrac{y-2}{-6}=\dfrac{z-3}{4}7x+5​=−6y−2​=4z−3​ intersect at the point R. The reflection f R in the xy – plane has coordinates:
  1. (A)(2, –4, –7)
  2. (B)(2, 4, 7)
  3. (C)(2, –4, 7)
  4. (D)(–2, 4, 7)

Correct answer: (A)

Step-by-step solution →
Q329·MathematicsSingle correctJEE Main 2019
Let A be a point on the line r⃗=(1−3μ)i^+(μ−1)j^+(2+5μ)k^\vec{r} = (1 - 3\mu)\hat{i} + (\mu - 1)\hat{j} + (2 + 5\mu)\hat{k}r=(1−3μ)i^+(μ−1)j^​+(2+5μ)k^ and B(3, 2, 6) be a point in the space. Then the value of μ\muμ for which the vector AB→\overrightarrow{AB}AB is parallel to the plane x−4y+3z=1x - 4y + 3z = 1x−4y+3z=1 is:
  1. (A)14\dfrac{1}{4}41​
  2. (B)18\dfrac{1}{8}81​
  3. (C)12\dfrac{1}{2}21​
  4. (D)−14-\dfrac{1}{4}−41​

Correct answer: (A)

Step-by-step solution →
Q330·MathematicsSingle correctJEE Main 2019
If the lines x=ay+b,z=cy+dx = ay + b, z = cy + dx=ay+b,z=cy+d and x=a′z+b′,y=c′z+d′x = a'z + b', y = c'z + d'x=a′z+b′,y=c′z+d′ are perpendicular, then:
  1. (A)cc′+a+a′=0cc' + a + a' = 0cc′+a+a′=0
  2. (B)aa′+c+c′=0aa' + c + c' = 0aa′+c+c′=0
  3. (C)ab′+bc′+1=0ab' + bc' + 1 = 0ab′+bc′+1=0
  4. (D)bb′+cc′+1=0bb' + cc' + 1 = 0bb′+cc′+1=0

Correct answer: (B)

Step-by-step solution →
Q331·MathematicsSingle correctJEE Main 2019
The plane through the intersection of the plane x+y+z=1x+y+z=1x+y+z=1 and 2x+3y−z+4=02x+3y-z+4=02x+3y−z+4=0 and parallel to y – axis also pass through the point:
  1. (A)(−3, 0, −1)(-3,\ 0,\ -1)(−3, 0, −1)
  2. (B)(−3, 1, 1)(-3,\ 1,\ 1)(−3, 1, 1)
  3. (C)(3, 3, −1)(3,\ 3,\ -1)(3, 3, −1)
  4. (D)(3, 2, 1)(3,\ 2,\ 1)(3, 2, 1)

Correct answer: (D)

Step-by-step solution →
Q332·MathematicsSingle correctJEE Main 2019
The equation of the plane containing the straight line x2=y3=z4\frac{x}{2} = \frac{y}{3} = \frac{z}{4}2x​=3y​=4z​ and perpendicular to the plane containing the straight lines x3=y4=z2\frac{x}{3} = \frac{y}{4} = \frac{z}{2}3x​=4y​=2z​ and x4=y2=z3\frac{x}{4} = \frac{y}{2} = \frac{z}{3}4x​=2y​=3z​ is:
  1. (A)x+2y−2z=0x + 2y - 2z = 0x+2y−2z=0
  2. (B)x−2y+z=0x - 2y + z = 0x−2y+z=0
  3. (C)5x+2y−4z=05x + 2y - 4z = 05x+2y−4z=0
  4. (D)3x+2y−3z=03x + 2y - 3z = 03x+2y−3z=0

Correct answer: (B)

Step-by-step solution →
Q333·MathematicsNumericalJEE Advanced 2018
Let P be a point in the first octant, whose image Q in the plane x+y=3x + y = 3x+y=3 (that is, the line segment PQ is perpendicular to the plane x+y=3x + y = 3x+y=3 and the mid-point of PQ lies in the plane x+y=3x + y = 3x+y=3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is ______ .

Correct answer: 8

Step-by-step solution →
Q334·MathematicsMultiple correctJEE Advanced 2018
Let P1:2x+y−z=3P_{1} : 2x + y - z = 3P1​:2x+y−z=3 and P2:x+2y+z=2P_{2} : x + 2y + z = 2P2​:x+2y+z=2 be two planes. Then, which of the following statement(s) is (are) TRUE ?
  1. (A)The line of intersection of P1P_{1}P1​ and P2P_{2}P2​ has direction ratios 1, 2, −1-1−1
  2. (B)The line 3x−49=1−3y9=z3\frac{3x-4}{9} = \frac{1-3y}{9} = \frac{z}{3}93x−4​=91−3y​=3z​ is perpendicular to the line of intersection of P1P_{1}P1​ and P2P_{2}P2​
  3. (C)The acute angle between P1P_{1}P1​ and P2P_{2}P2​ is 60∘60^{\circ}60∘
  4. (D)If P3P_{3}P3​ is the plane passing through the point (4,2,−2)(4, 2, -2)(4,2,−2) and perpendicular to the line of intersection of P1P_{1}P1​ and P2P_{2}P2​, then the distance of the point (2,1,1)(2, 1, 1)(2,1,1) from the plane P3P_{3}P3​ is 23\frac{2}{\sqrt{3}}3​2​

Correct answer: (C), (D)

Step-by-step solution →
Q335·MathematicsSingle correctJEE Advanced 2017
The equation of the plane passing through the point (1,1,1)(1, 1, 1)(1,1,1) and perpendicular to the planes 2x+y−2z=52x + y - 2z = 52x+y−2z=5 and 3x−6y−2z=73x - 6y - 2z = 73x−6y−2z=7, is
  1. (A)14x+2y−15z=114x + 2y - 15z = 114x+2y−15z=1
  2. (B)14x−2y+15z=2714x - 2y + 15z = 2714x−2y+15z=27
  3. (C)14x+2y+15z=3114x + 2y + 15z = 3114x+2y+15z=31
  4. (D)−14x+2y+15z=3-14x + 2y + 15z = 3−14x+2y+15z=3

Correct answer: (C)

Step-by-step solution →
Q336·MathematicsMultiple correctJEE Advanced 2016
Consider a pyramid OPQRS located in the first octant (x≥0x \geq 0x≥0, y≥0y \geq 0y≥0, z≥0z \geq 0z≥0) with O as origin, and OP and OR along the x-axis and the y-axis, respectively. The bases OPQR of the pyramid is a square with OP = 3. The point S is directly above the mid-point T of diagonal OQ such that TS = 3. Then
  1. (A)the acute angle between OQ and OS is π3\frac{\pi}{3}3π​
  2. (B)the equation of the plane containing the triangle OQS is x−y=0x - y = 0x−y=0
  3. (C)the length of the perpendicular from P to the plane containing the triangle OQS is 32\frac{3}{\sqrt{2}}2​3​
  4. (D)the perpendicular distance from O to the straight line containing RS is 152\sqrt{\frac{15}{2}}215​​

Correct answer: (B), (C), (D)

Step-by-step solution →
Q337·MathematicsSingle correctJEE Advanced 2016
Let P be the image of the point (3, 1, 7) with respect to the plane x−y+z=3x - y + z = 3x−y+z=3. Then the equation of the plane passing through P and containing the straight line x1=y2=z1\frac{x}{1} = \frac{y}{2} = \frac{z}{1}1x​=2y​=1z​ is
  1. (A)x+y−3z=0x + y - 3z = 0x+y−3z=0
  2. (B)3x+z=03x + z = 03x+z=0
  3. (C)x−4y+7z=0x - 4y + 7z = 0x−4y+7z=0
  4. (D)2x−y=02x - y = 02x−y=0

Correct answer: (C)

Step-by-step solution →
Q338·MathematicsMultiple correctJEE Advanced 2015
In R3\mathbb{R}^{3}R3, let LLL be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P1:x+2y−z+1=0P_{1} : x + 2y - z + 1 = 0P1​:x+2y−z+1=0 and P2:2x−y+z−1=0P_{2} : 2x - y + z - 1 = 0P2​:2x−y+z−1=0. Let MMM be the locus of the feet of the perpendiculars drawn from the points on LLL to the plane P1P_{1}P1​. Which of the following points lie(s) on MMM ?
  1. (A)(0,−56,−23)\left(0, -\dfrac{5}{6}, -\dfrac{2}{3}\right)(0,−65​,−32​)
  2. (B)(−16,−13,16)\left(-\dfrac{1}{6}, -\dfrac{1}{3}, \dfrac{1}{6}\right)(−61​,−31​,61​)
  3. (C)(−56,0,16)\left(-\dfrac{5}{6}, 0, \dfrac{1}{6}\right)(−65​,0,61​)
  4. (D)(−13,0,23)\left(-\dfrac{1}{3}, 0, \dfrac{2}{3}\right)(−31​,0,32​)

Correct answer: (A), (B)

Step-by-step solution →
Q339·MathematicsMultiple correctJEE Advanced 2015
In R3\mathbb{R}^{3}R3, consider the planes P1:y=0P_{1} : y = 0P1​:y=0 and P2:x+z=1P_{2} : x + z = 1P2​:x+z=1. Let P3P_{3}P3​ be a plane, different from P1P_{1}P1​ and P2P_{2}P2​, which passes through the intersection of P1P_{1}P1​ and P2P_{2}P2​. If the distance of the point (0,1,0)(0, 1, 0)(0,1,0) from P3P_{3}P3​ is 1 and the distance of a point (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) from P3P_{3}P3​ is 2, then which of the following relations is (are) true ?
  1. (A)2α+β+2γ+2=02\alpha + \beta + 2\gamma + 2 = 02α+β+2γ+2=0
  2. (B)2α−β+2γ+4=02\alpha - \beta + 2\gamma + 4 = 02α−β+2γ+4=0
  3. (C)2α+β−2γ−10=02\alpha + \beta - 2\gamma - 10 = 02α+β−2γ−10=0
  4. (D)2α−β+2γ−8=02\alpha - \beta + 2\gamma - 8 = 02α−β+2γ−8=0

Correct answer: (B), (D)

Step-by-step solution →
Q340·MathematicsMultiple correctJEE Advanced 2014
From a point P(λ,λ,λ)P(\lambda, \lambda, \lambda)P(λ,λ,λ), perpendiculars PQPQPQ and PRPRPR are drawn respectively on the lines y=x,z=1y = x, z = 1y=x,z=1 and y=−x,z=−1y = -x, z = -1y=−x,z=−1. If PPP is such that ∠QPR\angle QPR∠QPR is a right angle, then the possible value(s) of λ\lambdaλ is(are)
  1. (A)2\sqrt{2}2​
  2. (B)111
  3. (C)−1-1−1
  4. (D)−2-\sqrt{2}−2​

Correct answer: (C)

Step-by-step solution →
Q341·MathematicsSingle correctJEE Advanced 2013
Consider the lines L1:x−12=y−1=z+31L_{1} : \frac{x - 1}{2} = \frac{y}{-1} = \frac{z + 3}{1}L1​:2x−1​=−1y​=1z+3​, L2:x−41=y+31=z+32L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}L2​:1x−4​=1y+3​=2z+3​ and the planes P1:7x+y+2z=3P_{1} : 7x + y + 2z = 3P1​:7x+y+2z=3, P2:3x+5y−6z=4P_{2} : 3x + 5y - 6z = 4P2​:3x+5y−6z=4. Let ax+by+cz=dax + by + cz = dax+by+cz=d be the equation of the plane passing through the point of intersection of lines L1L_{1}L1​ and L2L_{2}L2​, and perpendicular to planes P1P_{1}P1​ and P2P_{2}P2​. Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.a=a =a=1.131313
Q.b=b =b=2.−3-3−3
R.c=c =c=3.111
S.d=d =d=4.−2-2−2
  1. (A)P-3, Q-2, R-4, S-1
  2. (B)P-1, Q-3, R-4, S-2
  3. (C)P-3, Q-2, R-1, S-4
  4. (D)P-2, Q-4, R-1, S-3

Correct answer: (A)

Step-by-step solution →
Q342·MathematicsMultiple correctJEE Advanced 2013
Two lines L1:x=5,y3−α=z−2L_{1} : x = 5, \frac{y}{3 - \alpha} = \frac{z}{-2}L1​:x=5,3−αy​=−2z​ and L2:x=α,y−1=z2−αL_{2} : x = \alpha, \frac{y}{-1} = \frac{z}{2 - \alpha}L2​:x=α,−1y​=2−αz​ are coplanar. Then α\alphaα can take value(s)
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)444

Correct answer: (A), (D)

Step-by-step solution →
Q343·MathematicsSingle correctJEE Advanced 2013
Perpendiculars are drawn from points on the line x+22=y+1−1=z3\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}2x+2​=−1y+1​=3z​ to the plane x+y+z=3x+y+z=3x+y+z=3. The feet of perpendiculars lie on the line
  1. (A)x5=y−18=z−2−13\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}5x​=8y−1​=−13z−2​
  2. (B)x2=y−13=z−2−5\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}2x​=3y−1​=−5z−2​
  3. (C)x4=y−13=z−2−7\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}4x​=3y−1​=−7z−2​
  4. (D)x2=y−1−7=z−25\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}2x​=−7y−1​=5z−2​

Correct answer: (D)

Step-by-step solution →
Q344·MathematicsMultiple correctJEE Advanced 2013
A line lll passing through the origin is perpendicular to the lines l1:(3+t)i^+(−1+2t)j^+(4+2t)k^l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}l1​:(3+t)i^+(−1+2t)j^​+(4+2t)k^, −∞<t<∞-\infty<t<\infty−∞<t<∞, l2:(3+2s)i^+(3+2s)j^+(2+s)k^l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}l2​:(3+2s)i^+(3+2s)j^​+(2+s)k^, −∞<s<∞-\infty<s<\infty−∞<s<∞. Then, the coordinate(s) of the point(s) on l2l_2l2​ at a distance of 17\sqrt{17}17​ from the point of intersection of lll and l1l_1l1​ is (are)
  1. (A)(73,73,53)(\frac{7}{3},\frac{7}{3},\frac{5}{3})(37​,37​,35​)
  2. (B)(−1,−1,0)(-1,-1,0)(−1,−1,0)
  3. (C)(1,1,1)(1,1,1)(1,1,1)
  4. (D)(79,79,89)(\frac{7}{9},\frac{7}{9},\frac{8}{9})(97​,97​,98​)

Correct answer: (B), (D)

Step-by-step solution →

Three Dimensional Geometry — frequently asked

How many questions from Three Dimensional Geometry appear in JEE?

Three Dimensional Geometry has appeared in 176 of the last 186 JEE Main and JEE Advanced papers — about 95% of them — contributing 344 questions in total across those papers.

Is Three Dimensional Geometry an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 95% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Three Dimensional Geometry questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

All 26 Mathematics chapters →

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