JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Three Dimensional Geometry

Let P be the plane such that it contains the straight line

\frac{x - 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{1}

and is perpendicular to the plane

x + 2 y + 3 z = 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 ?

(A)The equation of the plane P is $7 x - 5 y + z = - 10$ ✓ Correct
(B)The distance between the planes P and $P_{1}$ is 30
(C)The distance of the plane P from the origin is $23$
(D)The acute angle between the plane P and the plane $2 x + 2 y + z = 3$ is $cos^{- 1} (\frac{1}{3 3})$ ✓ Correct

Answer: (A), (D)

Solution

The line has direction vector

d = (2, 3, 1)

and passes through

(1, 3, - 2)

. The plane perpendicular to

x + 2 y + 3 z = 4

has normal perpendicular to

(1, 2, 3)

.

Normal to plane P:

n = d \times (1, 2, 3) = (7, - 5, 1)

.

Plane P through

(1, 3, - 2)

7 (x - 1) - 5 (y - 3) + (z + 2) = 0

gives

7 x - 5 y + z = - 10

. (A) is TRUE.

Plane

P_{1}

parallel to P through

(4, 2, 2)

7 x - 5 y + z = d

. Substituting:

d = 28 - 10 + 2 = 20

.

Distance between P and

P_{1}

\frac{∣20 - ( - 10 ) ∣}{49 + 25 + 1} = \frac{30}{75} = \frac{30}{5 3} = 23

. (B) is FALSE.

Distance from origin to P:

\frac{10}{75} = \frac{2 3}{3}

. (C) is FALSE.

Angle between P and

2 x + 2 y + z = 3

cos θ = \frac{∣14 - 10 + 1∣}{75 \cdot 3} = \frac{5}{15 3 / 3} = \frac{1}{3 3}

. Wait, let me recalculate:

cos θ = \frac{∣7 ( 2 ) - 5 ( 2 ) + 1 ( 1 ) ∣}{75 \cdot 9} = \frac{5}{5 3 \cdot 3} = \frac{1}{3 3}

. (D) is TRUE.

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · One or more correct options