3D Geometry- JEE Advanced Previous Year Questions with Solutions

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Why 3D Geometry Matters in JEE Advanced 

Three-dimensional geometry is one of the most consistently tested topics in JEE Advanced Mathematics. Questions from this chapter test spatial reasoning, vector fluency, and the ability to connect multiple concepts — lines, planes, distance formulas, and angle calculations — within a single problem.

According to the official JEE Advanced information brochure published by IIT (the organising institute rotates annually), Mathematics carries 108 marks in Paper 1 and Paper 2 combined. 3D Geometry alone has accounted for roughly 8–15% of the Mathematics score in most years between 2009 and 2023.

JEE Advanced Previous Year Questions 

 Download eSaral app for free study material and video tutorials. [esquestion] (A) Let $\mathrm{P}(3,2,6)$ be a point in space and $\mathrm{Q}$ be a point on the line $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}) .$ Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+3 z=1$ is - (A) $\frac{1}{4}$ (B) $-\frac{1}{4}$ (C) $\frac{1}{8}$ (D) $-\frac{1}{8}$ (B) A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals – (A) 1 (B) $\sqrt{2}$ (C) $\sqrt{3}$ (D) 2 (C) Let $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be points with integer coordinates satisfying the system of homogeneous equations $: 3 x-y-z=0 ;-3 x+z=0 ;-3 x+2 y+z=0 .$ Then the number of such points for which $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} \leq 100$ is #tag# [JEE 2009, 3+3+4] #sol# ( (a) $A ;(b) C ;(c) 7$ ) [/esquestion] [esquestion] (A) Equation of the plane containing the straight line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ (A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0 (C) x – 2y + z = 0 (D) 5x + 2y – 4z = 0 (B) If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=\alpha$ where $\alpha>0,$ is $5,$ then the foot of the perpendicular from $P$ to the plane is- (A) $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$ (B) $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$ (C) $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$ (D) $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$ (C) If the distance between the plane Ax – 2y + z = d and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6},$ then $|d|$ is (D) Match the statements in Column-I with the values in Column-II. #tag# [JEE 2010, 3+5+3+(2+2+2+2)] #sol# ((A) $\mathrm{C} ;(\mathrm{B}) \mathrm{A} ;(\mathrm{C}) 6 ;(\mathrm{D})(\mathrm{A}) \mathrm{t}(\mathrm{B}) \mathrm{p}, \mathrm{r}(\mathrm{C}) \mathrm{q}(\mathrm{D}) \mathrm{r}$ ) (a) Normal vector to the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is $\hat{n}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {3} & {4} & {2} \\ {4} & {2} & {3}\end{array}\right|=8 \hat{i}-\hat{j}-10 \hat{k}$ Let direction ratios of required plane be a, b, c. Now 8a – b – 10c = 0 and $2 \mathrm{a}+3 \mathrm{b}+4 \mathrm{c}=0\left(\because \text { plane contains the line } \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}\right)$ $\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$ $\cdot$ plane contains the line, which passes through origin, hence origin lies on a plane. $\Rightarrow$ equation of required plane is $x-2 y+z=0$ (b) $\quad \because \quad\left|\frac{1-4-2-\alpha}{3}\right|=5$ $\Rightarrow \alpha=10,-20$ $\Rightarrow \alpha=10 \because \alpha>0$ [/esquestion] [esquestion] (A) The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1,–1,4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2,1,4) to QR, then the length of the line segment PS is – (A) $\frac{1}{\sqrt{2}}$ (B) $\sqrt{2}$ (C) 2 (D) $2 \sqrt{2}$ (B) The equation of a plane passing through the line of intersection of the planes x + 2y $+3 \mathrm{z}=2$ and $\mathrm{x}-\mathrm{y}+\mathrm{z}=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is (A) $5 x-11 y+z=17$ (B) $\sqrt{2} x+y=3 \sqrt{2}-1$ (C) $x+y+z=\sqrt{3}$ (D) $x-\sqrt{2} y=1-\sqrt{2}$ (C) If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are) (A) y + 2z = –1 (B) y + z = –1 (C) y – z = –1 (D) y – 2z = –1 #tag# [JEE 2012, 3+3+4] #sol# ((a) $A ;(b) A ;(c) B, C$) (a) Line QR : $\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$ Any point on line QR : $(\lambda+2,4 \lambda+3, \lambda+5)$ $\therefore$ Point of intersection with plane : $5 \lambda+10-16 \lambda-12-\lambda-5=1$ $\Rightarrow \lambda=-\frac{2}{3}$ $\therefore \mathrm{P}\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$ [/esquestion] [esquestion] Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x$ $+\mathrm{y}+\mathrm{z}=3 .$ The feet of perpendiculars lie on the line (A) $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$ (B) $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$ (C) $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$ (D) $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$ #tag# [JEE-Advanced 2013, 2] #sol# (D) [/esquestion] [esquestion] A line $\ell$ passing through the origin is perpendicular to the lines $\ell_{1}:(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}},-\infty<\mathrm{t}<\infty$ $\ell_{2}:(3+2 s) \hat{\mathrm{i}}+(3+2 \mathrm{s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}},-\infty<\mathrm{s}<\infty$ Then, the coordinate(s) of the point(s) on $\ell_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $\ell$ and $\ell_{1}$ is (are) – (A) $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$ (B) (–1,–1,0) (C) (1,1,1) (D) $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$ #tag# [JEE-Advanced 2013, 4, (–1)] #sol# (B,D) [/esquestion] [esquestion] Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_{2}: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s) (A) 1              (B) 2             (C) 3                  (D) 4 #tag# [JEE-Advanced 2013, 3, (–1)] #sol# (A,D) $\mathrm{L}_{1}: \frac{\mathrm{x}-5}{0}=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}$ $\mathrm{L}_{2}: \frac{\mathrm{x}-\alpha}{0}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$ for lines to be coplanar $\left|\begin{array}{ccc}{5-\alpha} & {0} & {0} \\ {0} & {3-\alpha} & {-2} \\ {0} & {-1} & {2-\alpha}\end{array}\right|=0$ $\Rightarrow \quad(5-\alpha)((3-\alpha)(2-\alpha)-2)=0$ $\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right)=0$ $\Rightarrow \quad \alpha=1,4,5$ [/esquestion] [esquestion] Consider the lines $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+3}{1}, \mathrm{L}_{2}: \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{1}=\frac{\mathrm{z}+3}{2}$ and the planes $\mathrm{P}_{1}: 7 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=3, \mathrm{P}_{2}: 3 \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=4 .$ Let $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $\mathrm{L}_{2}$ and perpendicular to planes $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ Match List-I with List-II and select the correct answer using the code given below the lists. #tag# [JEE-Advanced 2013, 3, (–1)] #sol# (A) For point of intersection of $L_{1}$ and $L_{2}$ $\left\{\begin{array}{l}{2 \lambda+1=\mu+4} \\ {-\lambda=\mu-3} \\ {\lambda-3=2 \mu-3}\end{array}\right.$ $\Rightarrow \mu=1$ $\Rightarrow$ point of intersction is $(5,-2,-1)$ Now, vector normal to the plane is $\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {7} & {1} & {2} \\ {3} & {5} & {-6}\end{array}\right|$ $=-16(\hat{i}-3 \hat{j}-2 \hat{k})$ Let equation of required plane be $x-3 y-2 z=\alpha$ $\because$ it passes through $(5,-2,-1)$ $\therefore \alpha=13$ $\Rightarrow$ equation of plane is $x-3 y-2 z=13$ [/esquestion] [esquestion] From a point $\mathrm{P}(\lambda, \lambda, \lambda),$ perpendiculars $\mathrm{PQ}$ and $\mathrm{PR}$ are drawn respectively on the lines $\mathrm{y}=$ $\mathrm{x}, \mathrm{z}=1$ and $\mathrm{y}=-\mathrm{x}, \mathrm{z}=-1 .$ If $\mathrm{P}$ is such that $\angle \mathrm{QPR}$ is a right angle, then the possible value(s) of $\lambda$ is (are) (A) $\sqrt{2}$              (B) 1            (C) –1              (D) $-\sqrt{2}$ #tag# [JEE(Advanced)-2014, 3] #sol# (C) Line $\mathrm{L}_{1}$ given by $\mathrm{y}=\mathrm{x} ; \mathrm{z}=1$ can be expressed as $\mathrm{L}_{1}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}-1}{0}$ Similarly $\mathrm{L}_{2}(\mathrm{y}=-\mathrm{x} ; \mathrm{z}=-1)$ can be expressed as $\mathrm{L}_{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+1}{0}$ Let any point $\mathrm{Q}(\alpha, \alpha, 1)$ on $\mathrm{L}_{1}$ and $\mathrm{R}(\beta,-\beta,-1)$ on $\mathrm{L}_{2}$ Given that $\mathrm{PQ}$ is perpendicular to $\mathrm{L}_{1}$ $\Rightarrow(\lambda-\alpha) .1+(\lambda-\alpha) \cdot 1+(\lambda-1) \cdot 0=0 \Rightarrow \lambda=\alpha$ $\therefore \mathrm{Q}(\lambda, \lambda, 1)$ Similarly PR is perpendicular to L $_{2}$ $(\lambda-\beta) \cdot 1+(\lambda+\beta)(-1)+(\lambda+1) \cdot 0=0 \Rightarrow \beta=0$ $\therefore \mathrm{R}(0,0,-1)$ Now as given $\Rightarrow \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{PQ}}=0$ $0 . \lambda+0 . \lambda+(\lambda-1)(\lambda+1)=0$ $\lambda \neq 1$ as $\mathrm{P} \& \mathrm{Q}$ are different points $\Rightarrow \lambda=-1$ [/esquestion] [esquestion] In $\mathbb{D}^{3},$ consider the planes $P_{1}: y=0$ and $P_{2}: x+z=1 .$ Let $P_{3}$ be a plane, different from $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ which passes through the intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ If the distance of the point $(0,1,0)$ from $P_{3}$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_{3}$ is $2,$ then which of the following relations is (are) true? (A) $2 \alpha+\beta+2 \gamma+2=0$ (B) $2 \alpha-\beta+2 \gamma+4=0$ (C) $2 \alpha+\beta-2 \gamma-10=0$ (D) $2 \alpha-\beta+2 \gamma-8=0$ #tag# [JEE 2015, 4M, –2M] #sol# (B,D) [/esquestion] [esquestion] In $\square^{3},$ let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+$ $\mathrm{z}-1=0 .$ Let $\mathrm{M}$ be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1} .$ Which of the following points lie(s) on M? #tag# [JEE 2015, 4M, –2M] #sol# (A,B) Straight line 'L' is parallel to line of intersection of plane $\mathrm{P}_{1} \&$ plane $\mathrm{P}_{2}$ $\therefore$ Equation of line $^{\prime} \mathbf{L}^{\prime}$ is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$ $\frac{\alpha-\lambda}{1}=\frac{\beta+3 \lambda}{2}=\frac{\gamma+5 \lambda}{-1}=\mathrm{k}$ $\left.\begin{array}{l}{\alpha=\mathrm{k}+\lambda} \\ {\beta=2 \mathrm{k}-3 \lambda} \\ {\mathrm{y}=-\mathrm{k}-5 \lambda}\end{array}\right\}$ ...(1) satisfying in plane $\mathrm{P}_{1}$ $\mathrm{k}+\lambda+4 \mathrm{k}-6 \lambda+\mathrm{k}+5 \lambda+1=0$ $6 k=-1$ putting in ( 1) required locus is $\mathrm{x}=-\frac{1}{6}+\lambda$ $y=-\frac{1}{3}-3 \lambda$ $z=\frac{1}{6}-5 \lambda$ Now check the options. [/esquestion] [esquestion] Consider a pyramid OPQRS located in the first octant $(\mathrm{x} \geq 0, \mathrm{y} \geq 0, \mathrm{z} \geq 0)$ with $\mathrm{O}$ as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with $\mathrm{OP}=3 .$ The point $\mathrm{S}$ is directly above the mid-point $\mathrm{T}$ of diagonal OQ such that TS $=3 .$ Then- (A) the acute angle between $\mathrm{OQ}$ and $\mathrm{OS}$ is $\frac{\mathrm{K}}{3}$ (B) the equaiton of the plane containing the triangle $\mathrm{OQS}$ is $\mathrm{x}-\mathrm{y}=0$ (C) the length of the perpendicular from $P$ to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$ (D) the perpendicular distance from $\mathrm{O}$ to the straight line containing RS is $\sqrt{\frac{15}{2}}$ #tag# [JEE(Advanced) 2016] #sol# (B,C,D) [/esquestion] [esquestion] Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3 .$ Then the equation of the plane passing through $\mathrm{P}$ and containing the straight line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$ is (A) x + y – 3z = 0 (B) 3x + z = 0 (C) x – 4y + 7z = 0 (D) 2x – y = 0 #tag# [JEE(Advanced) 2016] #sol# (C) $\therefore x-4 y+7 z=0$ [/esquestion] [esquestion] The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is- (A) 14x + 2y + 15z = 31 (B) 14x + 2y – 15z = 1 (C) –14x + 2y + 15z = 3 (D) 14x – 2y + 15z = 27 #tag# [JEE(Advanced) 2017] #sol# (A) [/esquestion] [esquestion] Let $P_{1}: 2 x+y-z=3$ and $P_{2}: x+2 y+z=2$ be two planes. Then, which of the following statement(s) is (are) TRUE ? (A) The line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ has direction ratios $1,2,-1$ (B) The line $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$ (C) The acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is $60^{\circ}$ (D) If $P_{3}$ is the plane passing through the point $(4,2,-2)$ and perpendicular to the line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ then the distance of the point $(2,1,1)$ from the plane $\mathrm{P}_{2}$ is $\frac{2}{\sqrt{3}}$ #tag# [JEE(Advanced) 2018] #sol# (C,D) D.C. of line of intersection $(a, b, c)$ $\begin{aligned} \Rightarrow \quad & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}=0 \\ & \mathrm{a}+2 \mathrm{b}+\mathrm{c}=0 \end{aligned}$ $\frac{a}{1+2}=\frac{b}{-1-2}=\frac{c}{4-1}$ $\therefore \quad \mathrm{D} . \mathrm{C} .$ is $(1,-1,1)$ B) $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$ $\Rightarrow \quad \frac{x-4 / 3}{3}=\frac{y-1 / 3}{-3}=\frac{z}{3}$ $\Rightarrow \quad$ lines are parallel. (C) Acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}=\cos ^{-1}\left(\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{6} \sqrt{6}}\right)$ \[ =\cos ^{-1}\left(\frac{3}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ} \] (D) Plane is given by $(x-4)-(y-2)+(z+2)=0$ \[ \Rightarrow \quad x-y+z=0 \] Distance of $(2,1,1)$ from plane $=\frac{2-1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$ [/esquestion] [esquestion] Consider the cube in the first octant with sides $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ of length $1,$ along the x-axis, y-axis and z-axis, respectively, where $\mathrm{O}(0,0,0)$ is the origin. Let $\mathrm{S}\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT.If $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{SP}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{SQ}}, \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{SR}}$ and $\overrightarrow{\mathrm{t}}=\overrightarrow{\mathrm{ST}},$ then the value of $|(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{t}})|$ is #tag# [JEE(Advanced) 2018] #sol# 8 [/esquestion] [esquestion] Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + 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 #tag# #sol# 8 [/esquestion]

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