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(B) The equations of the planes are $P_{1}: x-y+2 z=2$ and $P_{2}: 2 x+y-z=2$.Let the line of intersection of planes $P_{1}$ and $P_{2}$ cut the $xy$-

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$119$

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(B) The equations of the planes are $P_{1}: x-y+2 z=2$ and $P_{2}: 2 x+y-z=2$.Let the line of intersection of planes $P_{1}$ and $P_{2}$ cut the $xy$-plane at point $Q$.Setting $z=0$ in both equations,we get $x-y=2$ and $2x+y=2$. Adding these,$3x=4 \Rightarrow x=\frac{4}{3}$,and $y=x-2 = \frac{4}{3}-2 = -\frac{2}{3}$.So,$Q = (\frac{4}{3}, -\frac{2}{3}, 0)$.The direction vector $\vec{a}$ of the line of intersection is the cross product of the normals $\vec{n}_{1} = \hat{i}-\hat{j}+2\hat{k}$ and $\vec{n}_{2} = 2\hat{i}+\hat{j}-\hat{k}$.$\vec{a} = \vec{n}_{1} 	imes \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(1-2) - \hat{j}(-1-4) + \hat{k}(1+2) = -\hat{i} + 5\hat{j} + 3\hat{k}$.The equation of the line $L$ is $\frac{x-4/3}{-1} = \frac{y+2/3}{5} = \frac{z}{3} = \lambda$.Any point $F$ on $L$ is $(-\lambda + 4/3, 5\lambda - 2/3, 3\lambda)$.Let $A = (1, 2, 0)$. The vector $\vec{AF} = (-\lambda + 4/3 - 1, 5\lambda - 2/3 - 2, 3\lambda - 0) = (-\lambda + 1/3, 5\lambda - 8/3, 3\lambda)$.Since $AF \perp L$,$\vec{AF} \cdot \vec{a} = 0$.$(-\lambda + 1/3)(-1) + (5\lambda - 8/3)(5) + (3\lambda)(3) = 0$.$\lambda - 1/3 + 25\lambda - 40/3 + 9\lambda = 0$.$35\lambda - 41/3 = 0 \Rightarrow 35\lambda = 41/3 \Rightarrow \lambda = 41/105$.The coordinates of the foot of the perpendicular $P(\alpha, \beta, \gamma)$ are $F$.$\alpha + \beta + \gamma = (-\lambda + 4/3) + (5\lambda - 2/3) + 3\lambda = 7\lambda + 2/3$.$35(\alpha + \beta + \gamma) = 35(7\lambda + 2/3) = 245\lambda + 70/3$.Substituting $\lambda = 41/105$: $245(41/105) + 70/3 = (49 	imes 41)/21 + 70/3 = (7 	imes 41)/3 + 70/3 = 287/3 + 70/3 = 357/3 = 119$.

Let $L$ be the line of intersection of planes $\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=2$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=2$. If $P(\alpha, \beta, \gamma)$ is the foot of perpendicular on $L$ from the point $(1,2,0)$,then the value of $35(\alpha+\beta+\gamma)$ is equal to :

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