Direction Ratios and Geometry of Intersecting Planes

To determine which statements are true, we analyze each option step-by-step:

1. Normal Vectors to Planes $P_1$ and $P_2$: The equations of the given planes are: $P_1: 2x + y - z = 3$ $P_2: x + 2y + z = 2$

The normal vectors to $P_1$ and $P_2$ are: $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$ $\vec{n}_2 = \hat{i} + 2\hat{j} + \hat{k}$

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2. Direction Ratios of the Line of Intersection (Evaluation of Option A): The direction vector $\vec{v}_L$ of the line of intersection of $P_1$ and $P_2$ is parallel to $\vec{n}_1 \times \vec{n}_2$:

$$\vec{v}_L = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix}$$

$\vec{v}_L = \hat{i}(1 - (-2)) - \hat{j}(2 - (-1)) + \hat{k}(4 - 1) = 3\hat{i} - 3\hat{j} + 3\hat{k} = 3(\hat{i} - \hat{j} + \hat{k})$

Thus, the direction ratios of the line of intersection are proportional to $\langle 1, -1, 1 \rangle$.
Since $\langle 1, 2, -1 \rangle$ is not proportional to $\langle 1, -1, 1 \rangle$, Option (A) is FALSE.

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3. Perpendicularity with the Given Line (Evaluation of Option B): The given line is: $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$

Rewriting it in standard form: $\frac{x - \frac{4}{3}}{3} = \frac{y - \frac{1}{3}}{-3} = \frac{z}{3}$

The direction vector of this line is $\vec{v} = 3\hat{i} - 3\hat{j} + 3\hat{k} = 3(\hat{i} - \hat{j} + \hat{k})$, which is parallel to $\vec{v}_L$.
Since $\vec{v} \cdot \vec{v}_L = 3(1) + (-3)(-1) + 3(1) = 9 \neq 0$, the line is parallel to the line of intersection, not perpendicular.
Therefore, Option (B) is FALSE.

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4. Angle Between the Planes (Evaluation of Option C): The angle $\theta$ between $P_1$ and $P_2$ is given by: $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} = \frac{|(2)(1) + (1)(2) + (-1)(1)|}{\sqrt{2^2 + 1^2 + (-1)^2} \sqrt{1^2 + 2^2 + 1^2}} = \frac{|2 + 2 - 1|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$

$\implies \theta = 60^\circ$

Therefore, the acute angle between $P_1$ and $P_2$ is $60^\circ$.
Thus, Option (C) is TRUE.

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5. Equation of Plane $P_3$ and Distance (Evaluation of Option D): The plane $P_3$ is perpendicular to the line of intersection of $P_1$ and $P_2$, which means the normal to $P_3$ is parallel to $\vec{v}_L = \hat{i} - \hat{j} + \hat{k}$.

Since $P_3$ passes through $(4, 2, -2)$, its equation is: $1(x - 4) - 1(y - 2) + 1(z - (-2)) = 0$ $x - y + z - 4 + 2 + 2 = 0 \implies x - y + z = 0$

The perpendicular distance $d$ from the point $(2, 1, 1)$ to $P_3$ is: $d = \frac{|2 - 1 + 1|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{2}{\sqrt{3}}$

Thus, Option (D) is TRUE.

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Correct Options: C, D