3D Geometry
390 Questions
Start JEE Mains Test
2021
Q251
JEE Mains
MCQ
14 Mar 2026
Let P be a plane lx + my + nz = 0 containing
the line, ${{1 - x} \over 1} = {{y + 4} \over 2} = {{z + 2} \over 3}$. 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 :
the line, ${{1 - x} \over 1} = {{y + 4} \over 2} = {{z + 2} \over 3}$. 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 :
A.
2
B.
3
C.
1.5
D.
4
2021
Q252
JEE Mains
MCQ
14 Mar 2026
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 :
A.
$\sqrt {41} $
B.
$\sqrt {55} $
C.
$\sqrt {31} $
D.
$\sqrt {66} $
2021
Q253
JEE Mains
MCQ
14 Mar 2026
If the mirror image of the point (1, 3, 5) with respect to the plane
4x $-$ 5y + 2z = 8 is ($\alpha$, $\beta$, $\gamma$), then 5($\alpha$ + $\beta$ + $\gamma$) equals :
4x $-$ 5y + 2z = 8 is ($\alpha$, $\beta$, $\gamma$), then 5($\alpha$ + $\beta$ + $\gamma$) equals :
A.
39
B.
41
C.
47
D.
43
2021
Q254
JEE Mains
MCQ
14 Mar 2026
Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P($\alpha$, $\beta$, $\gamma$) is the foot of perpendicular from (3, 2, 1) on L, then the
value of 21($\alpha$ + $\beta$ + $\gamma$) equals :
value of 21($\alpha$ + $\beta$ + $\gamma$) equals :
A.
102
B.
142
C.
136
D.
68
2021
Q255
JEE Mains
MCQ
14 Mar 2026
Consider the three planes
P1 : 3x + 15y + 21z = 9,
P2 : x $-$ 3y $-$ z = 5, and
P3 : 2x + 10y + 14z = 5
Then, which one of the following is true?
P1 : 3x + 15y + 21z = 9,
P2 : x $-$ 3y $-$ z = 5, and
P3 : 2x + 10y + 14z = 5
Then, which one of the following is true?
A.
P1 and P2 are parallel.
B.
P1, P2 and P3 all are parallel.
C.
P1 and P3 are parallel.
D.
P2 and P3 are parallel.
2021
Q256
JEE Mains
MCQ
14 Mar 2026
If (1, 5, 35), (7, 5, 5), (1, $\lambda$, 7) and (2$\lambda$, 1, 2) are coplanar, then the sum of all possible values of $\lambda$ is :
A.
$ - {{44} \over 5}$
B.
$ - {{39} \over 5}$
C.
${{44} \over 5}$
D.
${{39} \over 5}$
2021
Q257
JEE Mains
MCQ
14 Mar 2026
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 $\overrightarrow {OP} $ on this plane is of length :
A.
$\sqrt {{2 \over 7}} $
B.
$\sqrt {{2 \over 5}} $
C.
$\sqrt {{2 \over 3}} $
D.
$\sqrt {{2 \over 11}} $
2021
Q258
JEE Mains
MCQ
14 Mar 2026
The equation of the line through the point (0, 1, 2) and perpendicular to the line
${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}$ is :
${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}$ is :
A.
${x \over 3} = {{y - 1} \over { - 4}} = {{z - 2} \over 3}$
B.
${x \over 3} = {{y - 1} \over 4} = {{z - 2} \over { - 3}}$
C.
${x \over { - 3}} = {{y - 1} \over 4} = {{z - 2} \over 3}$
D.
${x \over 3} = {{y - 1} \over 4} = {{z - 2} \over 3}$
2021
Q259
JEE Mains
MCQ
14 Mar 2026
Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations l + m $-$ n = 0 and l2 + m2 $-$ n2 = 0. Then the value of sin4$\alpha$ + cos4$\alpha$ is :
A.
${{3 \over 8}}$
B.
${{3 \over 4}}$
C.
${{1 \over 2}}$
D.
${{5 \over 8}}$
2021
Q260
JEE Mains
MCQ
14 Mar 2026
Let a, b$ \in $R. If the mirror image of the point P(a, 6, 9) with respect to the line
${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$ is (20, b, $-$a$-$9), then | a + b |, is equal to :
${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$ is (20, b, $-$a$-$9), then | a + b |, is equal to :
A.
88
B.
90
C.
86
D.
84
2021
Q261
JEE Mains
MCQ
14 Mar 2026
The vector equation of the plane passing through the intersection
of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$, and the point (1, 0, 2) is :
of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$, and the point (1, 0, 2) is :
A.
$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = {7 \over 3}$
B.
$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = 7$
C.
$\overrightarrow r .\left( {3\widehat i + 7\widehat j + 3\widehat k} \right) = 7$
D.
$\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}$
2021
Q262
JEE Mains
MCQ
14 Mar 2026
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the
planes
3x + y - 2z = 5 and 2x - 5y - z = 7, is :
3x + y - 2z = 5 and 2x - 5y - z = 7, is :
A.
6x - 5y + 2z + 10 =0
B.
3x - 10y - 2z + 11 = 0
C.
6x - 5y - 2z - 2 = 0
D.
11x + y + 17z + 38 = 0
2021
Q263
JEE Mains
MCQ
14 Mar 2026
The distance of the point (1, 1, 9) from the point of intersection of the line
${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$
and the plane x + y + z = 17 is :
A.
$19\sqrt 2 $
B.
$2\sqrt {19} $
C.
38
D.
$\sqrt {38} $
2021
Q264
JEE Mains
Numerical
14 Mar 2026
Suppose, the line ${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ lies on the plane $x + 3y - 2z + \beta = 0$. Then $(\alpha + \beta )$ is equal to _______.
Correct Answer: 7
Explanation:
Given equation of line
${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ ...... (i)
and plane x + 3y $-$ 2z + $\beta$ = 0 ...... (ii)
Line (i) passes through (2, 2, $-$2)
which lies on plane (ii).
$\therefore$ 2 + 6 + 4 + $\beta$ = 0 $\Rightarrow$ $\beta$ = $-$ 12
Also, given line is perpendicular to normal of the plane
$\alpha$(1) $-$ 5(3) + 2($-$2) = 0 $\Rightarrow$ $\alpha$ = 19
$\therefore$ $\alpha$ + $\beta$ = 19 + (-12) = 19 - 12 = 7
2021
Q265
JEE Mains
Numerical
14 Mar 2026
The square of the distance of the point of intersection
of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.
of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.
Correct Answer: 61
Explanation:
${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $
$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$
for point of intersection of line & plane
$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$
$7\lambda = 7 \Rightarrow \lambda = 1$
point : (3, 5, 5)
(distance)2 = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$
$ = 16 + 36 + 9 = 61$
$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$
for point of intersection of line & plane
$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$
$7\lambda = 7 \Rightarrow \lambda = 1$
point : (3, 5, 5)
(distance)2 = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$
$ = 16 + 36 + 9 = 61$
2021
Q266
JEE Mains
Numerical
14 Mar 2026
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, $\gamma$) be a point of this plane. Then the square of the length of the line segment SR is ___________.
Correct Answer: 72
Explanation:
Since R(3, 5, $\gamma$) lies on the plane 2x $-$ y + z + 3 = 0.
Therefore, 6 $-$ 5 + $\gamma$ + 3 = 0
$\Rightarrow$ $\gamma$ = $-$4
Now,
dr's of line QS are 2, $-$1, 1
equation of line QS is
${{x - 1} \over 2} = {{y - 3} \over { - 1}} = {{z - 4} \over 1} = \lambda $ (say)
$ \Rightarrow F(2\lambda + 1, - \lambda + 3,\lambda + 4)$
F lies in the plane
$ \Rightarrow 2(2\lambda + 1) - ( - \lambda + 3) + (\lambda + 4) + 3$ = 0
$ \Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$
$ \Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = - 1$
$\Rightarrow$ F($-$1, 4, 3)
Since, F is mid-point of QS.
Therefore, coordinated of S are ($-$3, 5, 2).
So, SR = $\sqrt {36 + 0 + 36} = \sqrt {72} $
SR2 = 72.
2021
Q267
JEE Mains
Numerical
14 Mar 2026
Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)2, is equal to ___________.
Correct Answer: 96
Explanation:
Containing the line $\left| {\matrix{
{x + 1} & {y - 1} & {z - 3} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right| = 0$
$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$
$x - 2y + z = 0$
$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $
$P{Q^2} = 96$
$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$
$x - 2y + z = 0$
$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $
$P{Q^2} = 96$
2021
Q268
JEE Mains
Numerical
14 Mar 2026
Let the line L be the projection of the line ${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$ in the plane x $-$ 2y $-$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d2 is equal to _______________.
Correct Answer: 26
Explanation:
To find the projection let's find the foot of perpendicular from $(1,3$,
4) to plane $x-2 y-z=3$
$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $
So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.
Let us also find the intersection point of the plane and line
$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $
The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is
$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $
Now, let's find the distance from $O(0,0,6)$ to this line $L$.
Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,
$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $
$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $
So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.
Let us also find the intersection point of the plane and line
$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $
The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is
$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $
Now, let's find the distance from $O(0,0,6)$ to this line $L$.
Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,
$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $
2021
Q269
JEE Mains
Numerical
14 Mar 2026
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
Explanation:
$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$
$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$
Now, satisfying it in the given plane.
We get r = $-$2
so, required point of intersection is T(1, $-$2, 7).
Hence, PT = 7.
$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$
Now, satisfying it in the given plane.
We get r = $-$2
so, required point of intersection is T(1, $-$2, 7).
Hence, PT = 7.
2021
Q270
JEE Mains
Numerical
14 Mar 2026
Let a plane P pass through the point (3, 7, $-$7) and contain the line, ${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$. If distance of the plane P from the origin is d, then d2 is equal to ______________.
Correct Answer: 3
Explanation:
$\overrightarrow {BA} = (\widehat i + 4\widehat j - 5\widehat k)$
$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 3} & 2 & 1 \cr 1 & 4 & { - 5} \cr } } \right|$
$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$
a = 1, b = 1, c = 1
Plane is (x $-$ 2) + (y $-$ 3) + (z + 2) = 0
$ \Rightarrow $ x + y + z $-$ 3 = 0
$ \therefore $ d = $\sqrt 3 $ $\Rightarrow$ d2 = 3
$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 3} & 2 & 1 \cr 1 & 4 & { - 5} \cr } } \right|$
$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$
a = 1, b = 1, c = 1
Plane is (x $-$ 2) + (y $-$ 3) + (z + 2) = 0
$ \Rightarrow $ x + y + z $-$ 3 = 0
$ \therefore $ d = $\sqrt 3 $ $\Rightarrow$ d2 = 3
2021
Q271
JEE Mains
Numerical
14 Mar 2026
If the lines ${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$ and
${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ are co-planar, then the value of k is _____________.
${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ are co-planar, then the value of k is _____________.
Correct Answer: 1
Explanation:
$\left| {\matrix{
{k + 1} & 4 & 6 \cr
1 & 2 & 3 \cr
3 & 2 & 1 \cr
} } \right| = 0$
$ \Rightarrow $ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$
$ \Rightarrow $ $k = 1$
$ \Rightarrow $ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$
$ \Rightarrow $ $k = 1$
2021
Q272
JEE Mains
Numerical
14 Mar 2026
Let P be a plane passing through the points (1, 0, 1), (1, $-$2, 1) and (0, 1, $-$2). Let a vector $\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ be such that $\overrightarrow a $ is parallel to the plane P, perpendicular to $(\widehat i + 2\widehat j + 3\widehat k)$ and $\overrightarrow a \,.\,(\widehat i + \widehat j + 2\widehat k) = 2$, then ${(\alpha - \beta + \gamma )^2}$ equals ____________.
Correct Answer: 81
Explanation:
Equation of plane :
$\left| {\matrix{ {x - 1} & {y - 0} & {z - 1} \cr {1 - 1} & 2 & {1 - 1} \cr {1 - 0} & {0 - 1} & {1 + 2} \cr } } \right| = 0$
$ \Rightarrow 3x - z - 2 = 0$
$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0
$ \Rightarrow 3\alpha - 8 = 0$ ..... (1)
$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$
$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)
$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$
$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)
On solving 1, 2 & 3
$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3
So, ($\alpha$ $-$ $\beta$ + $\gamma$)2 = 81
$\left| {\matrix{ {x - 1} & {y - 0} & {z - 1} \cr {1 - 1} & 2 & {1 - 1} \cr {1 - 0} & {0 - 1} & {1 + 2} \cr } } \right| = 0$
$ \Rightarrow 3x - z - 2 = 0$
$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0
$ \Rightarrow 3\alpha - 8 = 0$ ..... (1)
$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$
$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)
$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$
$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)
On solving 1, 2 & 3
$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3
So, ($\alpha$ $-$ $\beta$ + $\gamma$)2 = 81
2021
Q273
JEE Mains
Numerical
14 Mar 2026
Let the mirror image of the point (1, 3, a) with respect to the plane $\overrightarrow r .\left( {2\widehat i - \widehat j + \widehat k} \right) - b = 0$ be ($-$3, 5, 2). Then, the value of | a + b | is equal to ____________.
Correct Answer: 1
Explanation:
Given equation of plane in vector form is $\overrightarrow r \,.\,(2\widehat i - \widehat j + \widehat k) - b = 0$
Its Cartesian form will be
$2x - y + z = b$ ...... (i)
$\because$ R is the mid-point of PQ.
$\therefore$ $R \equiv {{P + Q} \over 2} \Rightarrow R \equiv \left( { - 1,4,{{a + 2} \over 2}} \right)$
$\because$ R lies on the plane (i).
$\therefore$ $ - 2 - 4 + {{a + 2} \over 2} = b \Rightarrow a + 2 = 2b + 12$
$ \Rightarrow a = 2b + 10$ ....... (ii)
$\because$ Direction ratio's of QP is $(1 - ( - 3),3 - 5,a - 2)$
i.e. $(4, - 2,a - 2)$
and direction ratios of normal to the given plane are (2, $-$1, 1)
$\because$ n and QP are parallel.
$\therefore$ ${2 \over 4} = {{ - 1} \over { - 2}} = {1 \over {a - 2}}$
$\therefore$ $a - 2 = 2 \Rightarrow a = 4$
From Eq. (ii), b = $-$3
$\therefore$ $|a + b| = |4 - 3| = |1| = 1$
2021
Q274
JEE Mains
Numerical
14 Mar 2026
Let P be a plane containing the line ${{x - 1} \over 3} = {{y + 6} \over 4} = {{z + 5} \over 2}$ and parallel to the line ${{x - 1} \over 4} = {{y - 2} \over { - 3}} = {{z + 5} \over 7}$. If the point (1, $-$1, $\alpha$) lies on the plane P, then the value of |5$\alpha$| is equal to ____________.
Correct Answer: 38
Explanation:
Equation of required plane is $\left| {\matrix{ {x - 1} & {y + 6} & {z + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$
Since, (1, $-$1, $\alpha$) lies on it,
So, replace x by 1, y by ($-$1) and z and $\alpha$.
$\left| {\matrix{ 0 & 5 & {\alpha + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$
$ \Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$
$\therefore$ $\left| {5\alpha } \right| = \left| { - 38} \right| = 38$
2021
Q275
JEE Mains
Numerical
14 Mar 2026
Let the plane ax + by + cz + d = 0 bisect the line joining the points (4, $-$3, 1) and (2, 3, $-$5) at the right angles. If a, b, c, d are integers, then the
minimum value of (a2 + b2 + c2 + d2) is _________.
minimum value of (a2 + b2 + c2 + d2) is _________.
Correct Answer: 28
Explanation:
Normal of plane = $\overrightarrow {PQ} = - 2\widehat i + 6\widehat j - 6\widehat k$
a = $-$2, b = 6, c = $-$6
& equation of plane is
$-$2x + 6y $-$ 6z + d = 0
$ M(3,0, - 2)$ is the midpoint of the line which present on the plane
which satisfy the plane
$ \therefore $ d = $-$6
Now equation of plane is
$-$2x + 6y $-$ 6z $-$ 6 = 0
x $-$ 3y + 3z + 3 = 0
$ \Rightarrow $ (a2 + b2 + c2 + d2)min = 12 + 9 + 9 + 9 = 28
a = $-$2, b = 6, c = $-$6
& equation of plane is
$-$2x + 6y $-$ 6z + d = 0
$ M(3,0, - 2)$ is the midpoint of the line which present on the plane
which satisfy the plane
$ \therefore $ d = $-$6
Now equation of plane is
$-$2x + 6y $-$ 6z $-$ 6 = 0
x $-$ 3y + 3z + 3 = 0
$ \Rightarrow $ (a2 + b2 + c2 + d2)min = 12 + 9 + 9 + 9 = 28
2021
Q276
JEE Mains
Numerical
14 Mar 2026
The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $-$ d) = k(c $-$ a), then the positive value of k is :
Correct Answer: 4
Explanation:
The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0
$x - 2y + 2z + \lambda = 0$
Now given
$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$
$\left| {\lambda + 3} \right| = 3$
$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$
So planes are : $x - 2y + 2z - 6 = 0$
and $x - 2y + 2z = 0$
$b - d = - 2 + 6 = 4$
$c - a = 2 - 1 = 1$
$ \therefore $ $ {{b - d} \over {c - a}} = k$
$ \Rightarrow k = 4$
$x - 2y + 2z + \lambda = 0$
Now given
$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$
$\left| {\lambda + 3} \right| = 3$
$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$
So planes are : $x - 2y + 2z - 6 = 0$
and $x - 2y + 2z = 0$
$b - d = - 2 + 6 = 4$
$c - a = 2 - 1 = 1$
$ \therefore $ $ {{b - d} \over {c - a}} = k$
$ \Rightarrow k = 4$
2021
Q277
JEE Mains
Numerical
14 Mar 2026
Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx $-$ nz = 0 and x $-$ 2y + z = 0, equal to 9. If the locus of the point P is x2 + y2 + z2 = 9, then the value of l $-$ n is equal to _________.
Correct Answer: 0
Explanation:
Let point P is ($\alpha$, $\beta$, $\gamma$)
${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$
Locus is ${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$
${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$
Since its given that ${x^2} + {y^2} + {z^2} = 9$
After solving l = n,
then, l $-$ n = 0
${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$
Locus is ${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$
${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$
Since its given that ${x^2} + {y^2} + {z^2} = 9$
After solving l = n,
then, l $-$ n = 0
2021
Q278
JEE Mains
Numerical
14 Mar 2026
If the equation of the plane passing through the line of intersection of the planes 2x $-$ 7y + 4z $-$ 3 = 0, 3x $-$ 5y + 4z + 11 = 0 and the point ($-$2, 1, 3) is ax + by + cz $-$ 7 = 0, then the value of 2a + b + c $-$ 7 is ____________.
Correct Answer: 4
Explanation:
Equation of plane can be written using family of planes : P1 + $\lambda$P2 = 0
(2x $-$ 7y + 4z $-$ 3) + $\lambda$ (3x $-$ 5y + 4z + 11) = 0
It passes through ($-$2, 1, 3)
$ \therefore $ ($-$4 + 7 + 12 $-$ 3) + $\lambda$ ($-$6 $-$ 5 + 12 + 11) = 0
$-$2 + $\lambda$ (12) = 0
$\lambda$ = ${1 \over 6}$.
$ \therefore $ 12x $-$ 42y + 24z $-$ 18 + 3x $-$ 5y + 4z + 11 = 0
15x $-$ 47y + 28z $-$ 7 = 0
$ \therefore $ a = 15, b = $-$47, c = 28
$ \therefore $ 2a + b + c $-$ 7 = 30 $-$ 47 + 28 $-$ 7 = 4
(2x $-$ 7y + 4z $-$ 3) + $\lambda$ (3x $-$ 5y + 4z + 11) = 0
It passes through ($-$2, 1, 3)
$ \therefore $ ($-$4 + 7 + 12 $-$ 3) + $\lambda$ ($-$6 $-$ 5 + 12 + 11) = 0
$-$2 + $\lambda$ (12) = 0
$\lambda$ = ${1 \over 6}$.
$ \therefore $ 12x $-$ 42y + 24z $-$ 18 + 3x $-$ 5y + 4z + 11 = 0
15x $-$ 47y + 28z $-$ 7 = 0
$ \therefore $ a = 15, b = $-$47, c = 28
$ \therefore $ 2a + b + c $-$ 7 = 30 $-$ 47 + 28 $-$ 7 = 4
2021
Q279
JEE Mains
Numerical
14 Mar 2026
If the distance of the point (1, $-$2, 3) from the plane x + 2y $-$ 3z + 10 = 0 measured parallel to the line, ${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$ is $\sqrt {{7 \over 2}} $, then the value of |m| is equal to _________.
Correct Answer: 2
Explanation:
Given line L,
${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$
$ \Rightarrow $ ${{x - 1} \over 3} = {{y - 2} \over -m} = {{z + 3} \over 1}$
$ \therefore $ D.R of line = <3, -m, 1>
D.R of parallel line PQ will also be same.
$ \therefore $ Equation of line PQ,
${{x - 1} \over 3} = {{y + 2} \over { - m}} = {{z - 3} \over 1} = \lambda $
Pt. $Q(3\lambda + 1, - m\lambda - 2,\lambda + 3)$ lie on plane
$(3\lambda + 1) + 2( - m\lambda - 2) - 3(\lambda + 3) + 10 = 0$
$ \Rightarrow 3\lambda - 2m\lambda - 3\lambda + 1 - 4 - 9 + 10 = 0$
$ \Rightarrow - 2m\lambda = 2$
$m\lambda = - 1 \Rightarrow \lambda = - {1 \over m}$
$Q\left[ { - {3 \over m} + 1, - 1, - {1 \over m} + 3} \right]$
Given, $PQ = \sqrt {{7 \over 2}} $
$ \Rightarrow $ $\sqrt {{{\left( { - {3 \over m}} \right)}^2} + 1 + {{\left( { - {1 \over m}} \right)}^2}} = \sqrt {{7 \over 2}} $
$ \Rightarrow {{10 + {m^2}} \over {{m^2}}} = {7 \over 2}$
$ \Rightarrow 20 + 2{m^2} = 7{m^2}$
$ \Rightarrow $ ${m^2} = 4 \Rightarrow |m| = 2$
2021
Q280
JEE Mains
Numerical
14 Mar 2026
Let ($\lambda$, 2, 1) be a point on the plane which passes through the point (4, $-$2, 2). If the plane is perpendicular to the line joining the points ($-$2, $-$21, 29) and ($-$1, $-$16, 23), then ${\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$ is equal to __________.
Correct Answer: 8
Explanation:
$\overrightarrow {AB} \bot \overrightarrow {PQ} $
$\left[ {(4 - \lambda )\widehat i - 4\widehat j + \widehat k} \right].\left[ { + \widehat i + 5\widehat j - 6\widehat k} \right] = 0$
$4 - \lambda - 20 - 6 = 0$
$ \Rightarrow $ $\lambda $ = -22
Now, ${\lambda \over {11}} = - 2$
$ \Rightarrow {\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$
$ \Rightarrow 4 + 8 - 4 = 8$
2021
Q281
JEE Mains
Numerical
14 Mar 2026
A line 'l' passing through origin is perpendicular to the lines
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $\sqrt {17} $ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________.
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $\sqrt {17} $ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________.
Correct Answer: 44
Explanation:
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $ D.R. of ${l_1} = 1,2,2$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $ D.R. of ${l_2} = 2,2,1$
D.R. of l is $ \bot $ to l1 & k2
$ \therefore $ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$
$ \therefore $ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$
Solving l & l1
$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$
$ \Rightarrow 2\lambda = \mu + 3$
$ - 3\lambda = 2\mu - 1$
$2\lambda = 2\mu + 4$
$ \Rightarrow \mu + 3 = 2\mu + 4$
$\mu = - 1$
$\lambda = 1$
$P(2, - 3,2)$ {intersection point}
Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l2
Now, $PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $
$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$
$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$
$ \Rightarrow 9{v^2} + 28v + 20 = 0$
$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$
$ \Rightarrow (9v + 10)(v + 2) = 0$
$ \Rightarrow v = - 2$ (rejected), $ - {{10} \over 9}$ (accepted)
$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$
$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ \therefore $ $18(a + b + c)$
$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ = 44$
${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $ D.R. of ${l_1} = 1,2,2$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $ D.R. of ${l_2} = 2,2,1$
D.R. of l is $ \bot $ to l1 & k2
$ \therefore $ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$
$ \therefore $ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$
Solving l & l1
$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$
$ \Rightarrow 2\lambda = \mu + 3$
$ - 3\lambda = 2\mu - 1$
$2\lambda = 2\mu + 4$
$ \Rightarrow \mu + 3 = 2\mu + 4$
$\mu = - 1$
$\lambda = 1$
$P(2, - 3,2)$ {intersection point}
Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l2
Now, $PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $
$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$
$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$
$ \Rightarrow 9{v^2} + 28v + 20 = 0$
$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$
$ \Rightarrow (9v + 10)(v + 2) = 0$
$ \Rightarrow v = - 2$ (rejected), $ - {{10} \over 9}$ (accepted)
$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$
$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ \therefore $ $18(a + b + c)$
$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ = 44$
2021
Q282
JEE Mains
Numerical
14 Mar 2026
Let $\lambda$ be an integer. If the shortest distance between the lines
x $-$ $\lambda$ = 2y $-$ 1 = $-$2z and x = y + 2$\lambda$ = z $-$ $\lambda$ is ${{\sqrt 7 } \over {2\sqrt 2 }}$, then the value of | $\lambda$ | is _________.
x $-$ $\lambda$ = 2y $-$ 1 = $-$2z and x = y + 2$\lambda$ = z $-$ $\lambda$ is ${{\sqrt 7 } \over {2\sqrt 2 }}$, then the value of | $\lambda$ | is _________.
Correct Answer: 1
Explanation:
${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$
${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)
Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$
${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)
Point on line = $(0, - 2\lambda ,\lambda )$
Distance between skew lines $ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$
$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$
$\overline {\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|} $
$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$
$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$
$ \Rightarrow |\lambda | = 1$
${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)
Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$
${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)
Point on line = $(0, - 2\lambda ,\lambda )$
Distance between skew lines $ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$
$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$
$\overline {\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|} $
$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$
$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$
$ \Rightarrow |\lambda | = 1$
2020
Q283
JEE Mains
MCQ
14 Mar 2026
A plane P meets the coordinate axes at A, B
and C respectively. The centroid of $\Delta $ABC is
given to be (1, 1, 2). Then the equation of the
line through this centroid and perpendicular to
the plane P is :
A.
${{x - 1} \over 1} = {{y - 1} \over 1} = {{z - 2} \over 2}$
B.
${{x - 1} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$
C.
${{x - 1} \over 2} = {{y - 1} \over 2} = {{z - 2} \over 1}$
D.
${{x - 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}$
2020
Q284
JEE Mains
MCQ
14 Mar 2026
The shortest distance between the lines
${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :
${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :
A.
1
B.
${1 \over 2}$
C.
${1 \over {\sqrt 2 }}$
D.
${1 \over {\sqrt 3 }}$
2020
Q285
JEE Mains
MCQ
14 Mar 2026
If for some $\alpha $ $ \in $ R, the lines
L1 : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
L2 : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar,
then the line L2 passes through the point :
L1 : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
L2 : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar,
then the line L2 passes through the point :
A.
(10, 2, 2)
B.
(2, –10, –2)
C.
(10, –2, –2)
D.
(–2, 10, 2)
2020
Q286
JEE Mains
MCQ
14 Mar 2026
If (a, b, c) is the image of the point (1, 2, -3) in
the line ${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$, then a + b + c is :
the line ${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$, then a + b + c is :
A.
1
B.
2
C.
3
D.
-1
2020
Q287
JEE Mains
MCQ
14 Mar 2026
The distance of the point (1, –2, 3) from
the plane x – y + z = 5 measured parallel to
the line ${x \over 2} = {y \over 3} = {z \over { - 6}}$ is :
the plane x – y + z = 5 measured parallel to
the line ${x \over 2} = {y \over 3} = {z \over { - 6}}$ is :
A.
7
B.
1
C.
${1 \over 7}$
D.
${7 \over 5}$
2020
Q288
JEE Mains
MCQ
14 Mar 2026
The plane which bisects the line joining, the
points (4, –2, 3) and (2, 4, –1) at right angles
also passes through the point :
A.
(4, 0, 1)
B.
(0, –1, 1)
C.
(0, 1, –1)
D.
(4, 0, –1)
2020
Q289
JEE Mains
MCQ
14 Mar 2026
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 :
A.
x – 2y + z = 1
B.
x + 2y – z = 1
C.
x – y – 2y = 1
D.
2x + y – z = 1
2020
Q290
JEE Mains
MCQ
14 Mar 2026
A plane passing through the point (3, 1, 1)
contains two lines whose direction ratios are 1,
–2, 2 and 2, 3, –1 respectively. If this plane also
passes through the point ($\alpha $, –3, 5), then
$\alpha $ is
equal to:
A.
-10
B.
10
C.
5
D.
-5
2020
Q291
JEE Mains
MCQ
14 Mar 2026
The plane passing through the points (1, 2, 1),
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :
A.
(0, 6, –2)
B.
(–2, 0, 1)
C.
(0, –6, 2)
D.
(2, 0 –1)
2020
Q292
JEE Mains
MCQ
14 Mar 2026
The mirror image of the point (1, 2, 3) in a plane
is
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
A.
(1, –1, 1)
B.
(–1, –1, –1)
C.
(–1, –1, 1)
D.
(1, 1, 1)
2020
Q293
JEE Mains
MCQ
14 Mar 2026
The shortest distance between the lines
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
A.
3
B.
${7 \over 2}\sqrt {30} $
C.
$3\sqrt {30} $
D.
$2\sqrt {30} $
2020
Q294
JEE Mains
MCQ
14 Mar 2026
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 :
A.
(4, 3, 2)
B.
(6, 5, - 2)
C.
(3, 4, -2)
D.
(6, 5, 2)
2020
Q295
JEE Mains
Numerical
14 Mar 2026
If the equation of a plane P, passing through the intersection of the planes,
x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $ \in $ R, then the distance of the point (3, 2, -1) from the plane P is...........
x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $ \in $ R, then the distance of the point (3, 2, -1) from the plane P is...........
Correct Answer: 3
Explanation:
Equation of plane P is
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
2020
Q296
JEE Mains
Numerical
14 Mar 2026
Let a plane P contain two lines
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
Correct Answer: 5
Explanation:
Given lines,
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
2020
Q297
JEE Mains
Numerical
14 Mar 2026
If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
Correct Answer: 3
Explanation:
Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
2020
Q298
JEE Mains
Numerical
14 Mar 2026
The projection of the line segment joining the
points (1, –1, 3) and (2, –4, 11) on the line
joining the points (–1, 2, 3) and (3, –2, 10)
is ____________.
Correct Answer: 8
Explanation:
Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
2020
Q299
JEE Mains
Numerical
14 Mar 2026
If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($\alpha $, 7, 1)
is
$\left( {{5 \over 3},{7 \over 3},{{17} \over 3}} \right)$, then $\alpha $ is equal to______.
Correct Answer: 4
Explanation:
Direction Ratio of PQ are
= (${5 \over 3} - 1$, ${7 \over 3} - 0$, ${{17} \over 3} - 3$)
= (2, 7, 8)
Direction ratio of line QA are
= ($\alpha - {5 \over 3}$, $7 - {7 \over 3}$, 1 - ${{17} \over 3}$)
= (3$\alpha $ – 5, 14, –14)
PQ is perpendicular to line QA
$ \therefore $ $\overrightarrow {PQ} .\overrightarrow {QA} $ = 0
$ \Rightarrow $ 2(3$\alpha $ – 5) + 7.14 + (–14).8 = 0
$ \Rightarrow $ $\alpha $ = 4
2019
Q300
JEE Mains
MCQ
14 Mar 2026
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
A.
${1 \over 3}$
B.
${1 \over {\sqrt 3 }}$
C.
3
D.
${\sqrt 3 }$