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(C) The direction vectors of the two given lines are $\vec{v_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$ and $\vec{v_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.Si

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$\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

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(C) The direction vectors of the two given lines are $\vec{v_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$ and $\vec{v_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.Since the required line is perpendicular to both lines, its direction vector $\vec{v}$ must be parallel to the cross product $\vec{v_1} 	imes \vec{v_2}$.$\vec{v} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -16 & 7 \ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k}$.Dividing by the common factor $12$, we get the direction vector as $2\hat{i} + 3\hat{j} + 6\hat{k}$.The line passes through the point $(1, 2, -4)$, so its position vector is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.The vector equation of the line is $\vec{r} = \vec{a} + \lambda\vec{v} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$ is . . . . . . .

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