Let P_1: 2x + y - z = 3 and P_2: x + 2y + z = | vedclass

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(A) Let the direction ratios of the line of intersection be $a, b, c$.Since the line lies in both planes,it is perpendicular to the normals of both pl

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$C, D$

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(A) Let the direction ratios of the line of intersection be $a, b, c$.Since the line lies in both planes,it is perpendicular to the normals of both planes,$\vec{n_1} = (2, 1, -1)$ and $\vec{n_2} = (1, 2, 1)$.Thus,the direction vector is $\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1 + 2) - \hat{j}(2 + 1) + \hat{k}(4 - 1) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.Dividing by $3$,the direction ratios are $1, -1, 1$. Thus,$(A)$ is incorrect.For $(B)$,the line is $\frac{x - 4/3}{3} = \frac{y - 1/3}{-3} = \frac{z}{3}$,which has direction ratios $3, -3, 3$ or $1, -1, 1$. Since this is parallel to the line of intersection,it is not perpendicular. Thus,$(B)$ is incorrect.For $(C)$,$\cos 	heta = \frac{|(2)(1) + (1)(2) + (-1)(1)|}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+2^2+1^2}} = \frac{|2+2-1|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Thus,$	heta = 60^{\circ}$. $(C)$ is correct.For $(D)$,the normal to $P_3$ is $(1, -1, 1)$. The equation is $1(x-4) - 1(y-2) + 1(z+2) = 0 \Rightarrow x - y + z = 0$.The distance of $(2, 1, 1)$ from $x - y + z = 0$ is $\frac{|2 - 1 + 1|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{2}{\sqrt{3}}$. $(D)$ is correct.

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