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(C) The required line passes through the point $(3, 1, 2)$ and is perpendicular to the lines with direction vectors $\vec{v_1} = (1, 2, 3)$ and $\vec{

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$\frac{x-3}{2}=\frac{y-1}{-7}=\frac{z-2}{4}$

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(C) The required line passes through the point $(3, 1, 2)$ and is perpendicular to the lines with direction vectors $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.The direction vector $\vec{b}$ of the required line is given by the cross product of $\vec{v_1}$ and $\vec{v_2}$:$\vec{b} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -3 & 2 & 5 \end{vmatrix}$$\vec{b} = \hat{i}(2 	imes 5 - 3 	imes 2) - \hat{j}(1 	imes 5 - 3 	imes (-3)) + \hat{k}(1 	imes 2 - 2 	imes (-3))$$\vec{b} = \hat{i}(10 - 6) - \hat{j}(5 + 9) + \hat{k}(2 + 6) = 4\hat{i} - 14\hat{j} + 8\hat{k}$We can simplify the direction vector by dividing by $2$: $\vec{b'} = 2\hat{i} - 7\hat{j} + 4\hat{k}$.The equation of the line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the point $(3, 1, 2)$ and direction vector $(2, -7, 4)$,we get:$\frac{x-3}{2} = \frac{y-1}{-7} = \frac{z-2}{4}$.

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

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