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(C) The angle between two planes is equal to the angle between their normal vectors. First, we find the normal vector $\overrightarrow{n_1}$ of plane

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$\cos^{-1}\left(\frac{\sqrt{6}}{10}\right)$

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(C) The angle between two planes is equal to the angle between their normal vectors. First, we find the normal vector $\overrightarrow{n_1}$ of plane $\pi_1$. Since $\pi_1$ is perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$, its normal $\overrightarrow{n_1}$ is parallel to the cross product of the normals of these two planes, $\overrightarrow{n_A} = (1, 2, 3)$ and $\overrightarrow{n_B} = (1, 2, 2)$. $\overrightarrow{n_1} = \overrightarrow{n_A} 	imes \overrightarrow{n_B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(2-3) + \hat{k}(2-2) = -2\hat{i} + \hat{j} + 0\hat{k}$. Next, we find the normal vector $\overrightarrow{n_2}$ of plane $\pi_2$. The normal vector is the vector joining the point $(1, 3, 2)$ and its foot of the perpendicular $(-1, 2, -3)$. $\overrightarrow{n_2} = (-1-1)\hat{i} + (2-3)\hat{j} + (-3-2)\hat{k} = -2\hat{i} - \hat{j} - 5\hat{k}$. The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}| |\overrightarrow{n_2}|}$. $\overrightarrow{n_1} \cdot \overrightarrow{n_2} = (-2)(-2) + (1)(-1) + (0)(-5) = 4 - 1 + 0 = 3$. $|\overrightarrow{n_1}| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{5}$. $|\overrightarrow{n_2}| = \sqrt{(-2)^2 + (-1)^2 + (-5)^2} = \sqrt{4+1+25} = \sqrt{30}$. $\cos 	heta = \frac{3}{\sqrt{5} \cdot \sqrt{30}} = \frac{3}{\sqrt{150}} = \frac{3}{5\sqrt{6}} = \frac{\sqrt{6}}{10}$. Therefore, $	heta = \cos^{-1}\left(\frac{\sqrt{6}}{10}ight)$.

$\pi_1$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$. If $(-1, 2, -3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ onto a plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

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