JEE Advanced 2026 · Paper 1 · Mathematics Q6

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MathematicsMultiple correct+4 / -1Question 6 of 48

Let $P$ be the plane such that it contains the straight line $\dfrac{x-1}2=\dfrac{y-3}3=\dfrac{z+2}1$ and is perpendicular to the plane $x+2y+3z=4$ . Let $P_1$ be the plane which passes through the point $(4,2,2)$ and is parallel to $P$ .

Then which of the following statements is (are) TRUE ?

The equation of the plane

P

7x-5y+z=-10

The distance between the planes

P

and

P_1

is 30

The distance of the plane

P

from the origin is

2\sqrt3

The acute angle between the plane

P

and the plane

2x+2y+z=3

\cos^{-1}\left(\dfrac1{3\sqrt3}\right)

Answer and worked solution

Official answer: A, D

Plane $P$ . Its normal is perpendicular to the line direction $(2,3,1)$ and to $(1,2,3)$ :
$\vec n=(2,3,1)\times(1,2,3)=(7,-5,1).$
Through $(1,3,-2)$ : $7(x-1)-5(y-3)+(z+2)=0$ , i.e. $7x-5y+z=-10$ . (A) true.

Plane $P_1$ . Through $(4,2,2)$ : $7x-5y+z=20$ . Distance between the planes:
$\frac{|20-(-10)|}{\sqrt{75}}=\frac{30}{5\sqrt3}=2\sqrt3.$ (B) false.

(C) Distance of $P$ from the origin: $\dfrac{10}{5\sqrt3}=\dfrac2{\sqrt3}$ . False.

(D) $\cos\theta=\dfrac{|14-10+1|}{5\sqrt3\cdot3}=\dfrac1{3\sqrt3}$ . True.

Topics:Equation of a plane Angles and distances Cross product

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JEE Advanced 2026 · Paper 1, Mathematics question 6