Q22 / 48 · JEE Advanced 2026 (Paper 1) · Mathematics
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?
Then which of the following statements is (are) TRUE?
- Correct answerAThe equation of the plane $P$ is $7x - 5y + z = -10$
- BThe distance between the planes $P$ and $P_1$ is $30$
- CThe distance of the plane $P$ from the origin is $2\sqrt{3}$
- Correct answerDThe acute angle between the plane $P$ and the plane $2x + 2y + z = 3$ is $\cos^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)$
Solution
Let the normal vector to the plane $P$ be $\vec{n}$.
Since the plane $P$ contains the line $\dfrac{x-1}{2} = \dfrac{y-3}{3} = \dfrac{z+2}{1}$, its normal vector $\vec{n}$ is perpendicular to the line's direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Since $P$ is perpendicular to the plane $x + 2y + 3z = 4$, $\vec{n}$ is perpendicular to its normal vector $\vec{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{n} = (2\hat{i} + 3\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i}(9-2) - \hat{j}(6-1) + \hat{k}(4-3) = 7\hat{i} - 5\hat{j} + \hat{k}$
Plane $P$ passes through the point $(1, 3, -2)$ which lies on the given line.
Equation of plane $P$ is $7(x-1) - 5(y-3) + 1(z+2) = 0 \Rightarrow 7x - 5y + z = -10$
Statement (1) is TRUE.
The plane $P_1$ is parallel to $P$ and passes through $(4, 2, 2)$.
Equation of plane $P_1$ is $7(x-4) - 5(y-2) + 1(z-2) = 0 \Rightarrow 7x - 5y + z = 20$
Distance between parallel planes $P$ and $P_1$ is $d = \dfrac{|20 - (-10)|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \dfrac{30}{\sqrt{75}} = 2\sqrt{3}$. Statement (2) is FALSE.
Distance of plane $P$ from the origin is $d_0 = \dfrac{|-10|}{\sqrt{75}} = \dfrac{2}{\sqrt{3}}$. Statement (3) is FALSE.
Angle $\theta$ between the planes $7x - 5y + z + 10 = 0$ and $2x + 2y + z - 3 = 0$ is given by:
$\cos\theta = \dfrac{|7(2) + (-5)(2) + 1(1)|}{\sqrt{7^2+(-5)^2+1^2} \sqrt{2^2+2^2+1^2}} = \dfrac{5}{\sqrt{75} \sqrt{9}} = \dfrac{5}{15\sqrt{3}} = \dfrac{1}{3\sqrt{3}}$
$\theta = \cos^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)$. Statement (4) is TRUE.
Since the plane $P$ contains the line $\dfrac{x-1}{2} = \dfrac{y-3}{3} = \dfrac{z+2}{1}$, its normal vector $\vec{n}$ is perpendicular to the line's direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Since $P$ is perpendicular to the plane $x + 2y + 3z = 4$, $\vec{n}$ is perpendicular to its normal vector $\vec{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{n} = (2\hat{i} + 3\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i}(9-2) - \hat{j}(6-1) + \hat{k}(4-3) = 7\hat{i} - 5\hat{j} + \hat{k}$
Plane $P$ passes through the point $(1, 3, -2)$ which lies on the given line.
Equation of plane $P$ is $7(x-1) - 5(y-3) + 1(z+2) = 0 \Rightarrow 7x - 5y + z = -10$
Statement (1) is TRUE.
The plane $P_1$ is parallel to $P$ and passes through $(4, 2, 2)$.
Equation of plane $P_1$ is $7(x-4) - 5(y-2) + 1(z-2) = 0 \Rightarrow 7x - 5y + z = 20$
Distance between parallel planes $P$ and $P_1$ is $d = \dfrac{|20 - (-10)|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \dfrac{30}{\sqrt{75}} = 2\sqrt{3}$. Statement (2) is FALSE.
Distance of plane $P$ from the origin is $d_0 = \dfrac{|-10|}{\sqrt{75}} = \dfrac{2}{\sqrt{3}}$. Statement (3) is FALSE.
Angle $\theta$ between the planes $7x - 5y + z + 10 = 0$ and $2x + 2y + z - 3 = 0$ is given by:
$\cos\theta = \dfrac{|7(2) + (-5)(2) + 1(1)|}{\sqrt{7^2+(-5)^2+1^2} \sqrt{2^2+2^2+1^2}} = \dfrac{5}{\sqrt{75} \sqrt{9}} = \dfrac{5}{15\sqrt{3}} = \dfrac{1}{3\sqrt{3}}$
$\theta = \cos^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)$. Statement (4) is TRUE.
Hence, the correct answer is options: 'A', 'D'.