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(D) The direction ratio of the line is given by the cross product of the normals to the two planes:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat

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$5$

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(D) The direction ratio of the line is given by the cross product of the normals to the two planes:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 0 & 3 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(-1-0) + \hat{k}(3-0) = -5\hat{i} + \hat{j} + 3\hat{k}$.The equation of the line passing through $(3, 2, 1)$ with direction vector $\vec{v} = \langle -5, 1, 3 angle$ is $\frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3} = \lambda$.Any point $M$ on the line is given by $M(-5\lambda+3, \lambda+2, 3\lambda+1)$.Let $P = (1, 9, 7)$. The vector $\vec{PM} = \langle -5\lambda+3-1, \lambda+2-9, 3\lambda+1-7 angle = \langle -5\lambda+2, \lambda-7, 3\lambda-6 angle$.Since $\vec{PM}$ is perpendicular to the line,$\vec{PM} \cdot \langle -5, 1, 3 angle = 0$.$-5(-5\lambda+2) + 1(\lambda-7) + 3(3\lambda-6) = 0$.$25\lambda - 10 + \lambda - 7 + 9\lambda - 18 = 0$.$35\lambda - 35 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ into the coordinates of $M$,we get $M = (-5(1)+3, 1+2, 3(1)+1) = (-2, 3, 4)$.Thus,$(\alpha, \beta, \gamma) = (-2, 3, 4)$.Therefore,$\alpha+\beta+\gamma = -2+3+4 = 5$.

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

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