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(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $

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$\frac{x-2}{-5} = \frac{y-3}{-4} = \frac{z-1}{3}$

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(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $\vec{N}_2 = \hat{i} + \hat{j} + 3\hat{k}$.The direction vector $\vec{b}$ of the line of intersection is given by the cross product of the normals:$\vec{b} = \vec{N}_1 	imes \vec{N}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & -1 \ 1 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - (-1)) - \hat{j}(3 - (-1)) + \hat{k}(1 - (-2)) = -5\hat{i} - 4\hat{j} + 3\hat{k}$.The line passes through $(2, 3, 1)$ and is parallel to $\vec{b} = -5\hat{i} - 4\hat{j} + 3\hat{k}$.Thus,the equation of the line is $\frac{x-2}{-5} = \frac{y-3}{-4} = \frac{z-1}{3}$.

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

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