Equation of line passing through the point \( (2,3,1) \) and parallel to the line of intersection of the
plane \( x-2 y-z+5=0 \) and \( x+y+3 z=6 \) is

Given equation of planes,
\[
\begin{array}{l}
P_{1}: x-2 y-z+5=0 \rightarrow(1) \\
P_{2}: x+y+3 z=6 \rightarrow(2)
\end{array}
\]
and point \( P(2,3,1) \).
Normal vector of Eq. (1) is given by,
\[
\vec{N}_{1}=\hat{i}-2 \hat{j}+\hat{k} \rightarrow(3)
\]
Similarly, normal vector of Eq. (2) is given by
\[
\vec{N}_{2}=\hat{i}+\hat{j}+3 \hat{k} \rightarrow(4)
\]
Vector perpendicular to the normal to the plans are:
\[
\begin{aligned}
\vec{b} &=\vec{N}_{1} \times \vec{N}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 1 \\
1 & 1 & 3
\end{array}\right| \\
&=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}
\end{aligned}
\]
So, required equation of line passing through the point \( (2,3,1) \) is,
\[
\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}
\]