JEE Advanced 2026 — Paper 1 — MATH-II solution

JEE Advanced2026Paper 1MATH-II

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 $7 x - 5 y + z = - 10$
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 $cos^{- 1} (\frac{1}{3 3})$

Solution

Normal of plane P: P contains the line with direction $d = ⟨ 2, 3, 1 ⟩$ and is perpendicular to $x + 2 y + 3 z = 4$ whose normal is $m = ⟨ 1, 2, 3 ⟩$ . So $n = d \times m$ :

$n = \hat{i} \hat{j} \hat{k} 231123 = 7 \hat{i} - 5 \hat{j} + \hat{k}$

(A) Using the point $(1, 3, - 2)$ on the given line: $7 (x - 1) - 5 (y - 3) + (z + 2) = 0 \Rightarrow 7 x - 5 y + z = - 10$ . TRUE.

(B) $P_{1}$ has the form $7 x - 5 y + z = d$ . Substituting $(4, 2, 2)$ gives $d = 20$ . Distance between parallel planes $= \frac{∣20 - ( - 10 ) ∣}{49 + 25 + 1} = \frac{30}{75} = \frac{30}{5 3} = 23$ , not $30$ . FALSE.

(C) Distance from origin to P $= \frac{∣ - 10 ∣}{75} = \frac{10}{5 3} = \frac{2 3}{3}$ , not $23$ . FALSE.

(D) With $n_{2} = ⟨ 2, 2, 1 ⟩$ : $cos θ = \frac{14 - 10 + 1}{75 \cdot 3} = \frac{5}{15 3} = \frac{1}{3 3}$ . TRUE.

$∴$ The correct options are (A) and (D).

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