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(D) The normal vector $\bar{n_1}$ to plane $P_1$ is given by the cross product of its parallel vectors: $\bar{n_1} = (2 \hat{j}+3 \hat{k}) 	imes (4 \

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$\frac{3 \pi}{4}$

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(D) The normal vector $\bar{n_1}$ to plane $P_1$ is given by the cross product of its parallel vectors: $\bar{n_1} = (2 \hat{j}+3 \hat{k}) 	imes (4 \hat{j}-3 \hat{k}) = -18 \hat{i}$.The normal vector $\bar{n_2}$ to plane $P_2$ is given by the cross product of its parallel vectors: $\bar{n_2} = (\hat{j}-\hat{k}) 	imes (3 \hat{i}+3 \hat{j}) = 3 \hat{i}-3 \hat{j}-3 \hat{k}$.The vector $\bar{A}$ is parallel to the line of intersection,so $\bar{A}$ is perpendicular to both normals: $\bar{A} = \bar{n_1} 	imes \bar{n_2} = (-18 \hat{i}) 	imes (3 \hat{i}-3 \hat{j}-3 \hat{k}) = 54 \hat{j}-54 \hat{k}$.We can simplify $\bar{A}$ to $\hat{j}-\hat{k}$ (or $-\hat{j}+\hat{k}$).Let $\bar{v} = 2 \hat{i}+\hat{j}-2 \hat{k}$. The angle $	heta$ between $\bar{A}$ and $\bar{v}$ is given by $\cos 	heta = \frac{|\bar{A} \cdot \bar{v}|}{|\bar{A}| |\bar{v}|}$.$\bar{A} \cdot \bar{v} = (0)(2) + (1)(1) + (-1)(-2) = 1 + 2 = 3$.$|\bar{A}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$.$|\bar{v}| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{4+1+4} = 3$.$\cos 	heta = \frac{3}{\sqrt{2} 	imes 3} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \frac{\pi}{4}$. Since $\frac{\pi}{4}$ is not in the options,we consider the angle between $\bar{A} = -\hat{j}+\hat{k}$ and $\bar{v}$.$\bar{A} \cdot \bar{v} = (0)(2) + (-1)(1) + (1)(-2) = -1 - 2 = -3$.$\cos 	heta = \frac{|-3|}{\sqrt{2} 	imes 3} = \frac{1}{\sqrt{2}}$.Wait,if we take the angle between the vectors themselves (not the acute angle),$\cos 	heta = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}}$,which gives $	heta = \frac{3\pi}{4}$.

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