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(D) Let the direction ratios of the line of intersection of the planes $x-y=0$ and $2x+y+z=0$ be $a_1, b_1, c_1$. Since the line lies in both planes,i

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$90$

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(D) Let the direction ratios of the line of intersection of the planes $x-y=0$ and $2x+y+z=0$ be $a_1, b_1, c_1$. Since the line lies in both planes,it is perpendicular to the normals of both planes. The normals are $\vec{n_1} = (1, -1, 0)$ and $\vec{n_2} = (2, 1, 1)$.The direction vector $\vec{v_1} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1+2) = -\hat{i} - \hat{j} + 3\hat{k}$.Thus,the direction ratios are $(a_1, b_1, c_1) = (-1, -1, 3)$,which is equivalent to $(1, 1, -3)$.Similarly,for the planes $2x-z=0$ and $x+y-3z=0$,the normals are $\vec{n_3} = (2, 0, -1)$ and $\vec{n_4} = (1, 1, -3)$.The direction vector $\vec{v_2} = \vec{n_3} 	imes \vec{n_4} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & -1 \ 1 & 1 & -3 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(-6+1) + \hat{k}(2-0) = \hat{i} + 5\hat{j} + 2\hat{k}$.Wait,re-calculating $\vec{v_2}$: $\hat{i}(0 - (-1)) - \hat{j}(-6 - (-1)) + \hat{k}(2 - 0) = \hat{i} + 5\hat{j} + 2\hat{k}$.Let us re-calculate the cross product for the second set: $\vec{n_3} = (2, 0, -1)$,$\vec{n_4} = (1, 1, -3)$. $\vec{v_2} = (0(-3) - (-1)(1), -((-1)(1) - 2(-3)), 2(1) - 0(1)) = (1, -5, 2)$.Now,the angle $	heta$ between $\vec{v_1} = (1, 1, -3)$ and $\vec{v_2} = (1, -5, 2)$ is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (1)(1) + (1)(-5) + (-3)(2) = 1 - 5 - 6 = -10$.$\cos 	heta = \frac{|-10|}{\sqrt{1^2+1^2+(-3)^2} \sqrt{1^2+(-5)^2+2^2}} = \frac{10}{\sqrt{11} \sqrt{30}}$.Re-checking the original problem values: $x-y=0, 2x+y+z=0 \Rightarrow \vec{v_1} = (1, 1, -3)$. $2x-z=0, x+y-3z=0 \Rightarrow \vec{v_2} = (1, 5, 2)$.$\vec{v_1} \cdot \vec{v_2} = 1 + 5 - 6 = 0$. Since the dot product is $0$,the angle is $90^{\circ}$.Hence,option $(d)$ is correct.

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)

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