Properties and Equations of Parallel and Perpendicular Planes

To determine which statements are correct, we analyze each step based on the given geometric conditions.

Step 1: Finding the Equation of Plane $P$

The straight line given is: $\frac{x-1}{2} = \frac{y-3}{3} = \frac{z+2}{1}$

From this equation, we can identify:

• A point lying on the line (and hence on plane $P$): $A(1, 3, -2)$
• The direction vector of the line: $\mathbf{v} = 2\hat{i} + 3\hat{j} + \hat{k}$

The normal vector to the plane $x + 2y + 3z = 4$ is: $\mathbf{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$

Since the plane $P$ contains the line and is perpendicular to the plane $x + 2y + 3z = 4$, its normal vector $\mathbf{n}_P$ must be perpendicular to both $\mathbf{v}$ and $\mathbf{n}_1$. Thus, $\mathbf{n}_P$ is parallel to $\mathbf{v} \times \mathbf{n}_1$:

$\mathbf{n}_P = \mathbf{v} \times \mathbf{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

$\mathbf{n}_P = \hat{i}(9 - 2) - \hat{j}(6 - 1) + \hat{k}(4 - 3) = 7\hat{i} - 5\hat{j} + \hat{k}$

The equation of the plane $P$ passing through point $A(1, 3, -2)$ is: $7(x - 1) - 5(y - 3) + 1(z + 2) = 0$ $7x - 7 - 5y + 15 + z + 2 = 0$ $7x - 5y + z + 10 = 0 \implies 7x - 5y + z = -10$

Thus, Statement (A) is TRUE.

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Step 2: Distance Between the Planes $P$ and $P_1$

The plane $P_1$ is parallel to $P$ and passes through the point $(4, 2, 2)$. The equation of plane $P_1$ is: $7(x - 4) - 5(y - 2) + 1(z - 2) = 0$ $7x - 28 - 5y + 10 + z - 2 = 0$ $7x - 5y + z - 20 = 0$

The perpendicular distance $d$ between the parallel planes $P: 7x - 5y + z + 10 = 0$ and $P_1: 7x - 5y + z - 20 = 0$ is given by: $d = \frac{|10 - (-20)|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \frac{30}{\sqrt{49 + 25 + 1}} = \frac{30}{\sqrt{75}} = \frac{30}{5\sqrt{3}} = 2\sqrt{3}$

Thus, Statement (B) is FALSE (the distance is $2\sqrt{3}$, not $30$).

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Step 3: Distance of Plane $P$ from the Origin

The distance $d_0$ from the origin $(0,0,0)$ to the plane $P: 7x - 5y + z + 10 = 0$ is: $d_0 = \frac{|10|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \frac{10}{\sqrt{75}} = \frac{10}{5\sqrt{3}} = \frac{2}{\sqrt{3}}$

Thus, Statement (C) is FALSE (the distance is $\frac{2}{\sqrt{3}}$, not $2\sqrt{3}$).

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Step 4: Acute Angle Between Plane $P$ and $2x + 2y + z = 3$

The normal vector to plane $P$ is $\mathbf{n}_P = 7\hat{i} - 5\hat{j} + \hat{k}$. The normal vector to the plane $2x + 2y + z = 3$ is $\mathbf{n}_2 = 2\hat{i} + 2\hat{j} + \hat{k}$.

The cosine of the acute angle $\theta$ between the two planes is given by: $\cos\theta = \frac{|\mathbf{n}_P \cdot \mathbf{n}_2|}{|\mathbf{n}_P| |\mathbf{n}_2|}$

Calculating the dot product and magnitudes: $\mathbf{n}_P \cdot \mathbf{n}_2 = (7)(2) + (-5)(2) + (1)(1) = 14 - 10 + 1 = 5$ $|\mathbf{n}_P| = \sqrt{75} = 5\sqrt{3}$ $|\mathbf{n}_2| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$

Substituting these values: $\cos\theta = \frac{5}{(5\sqrt{3})(3)} = \frac{1}{3\sqrt{3}}$ $\theta = \cos^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Thus, Statement (D) is TRUE.

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Conclusion

The correct options are A and D.