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(C) The required line passes through the point $A(1, 2, 3)$,so its position vector is $\overline{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.Since the line is

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$\overline{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-4\hat{i} - 7\hat{j} - 13\hat{k})$

Vedclass · Answer

(C) The required line passes through the point $A(1, 2, 3)$,so its position vector is $\overline{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.Since the line is perpendicular to the lines with direction vectors $\overline{b_1} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ and $\overline{b_2} = 2\hat{i} - 3\hat{j} + \hat{k}$,the direction vector $\overline{b}$ of the required line is given by the cross product $\overline{b_1} 	imes \overline{b_2}$.$\overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & -2 \ 2 & -3 & 1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(3 - (-4)) + \hat{k}(-9 - 4) = -4\hat{i} - 7\hat{j} - 13\hat{k}$.The equation of the line is $\overline{r} = \overline{a} + \lambda\overline{b}$,which is $\overline{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-4\hat{i} - 7\hat{j} - 13\hat{k})$.

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

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