Find an equation of the line of intersection of the two planes $$x-2y+z=-1\quad \hbox{and}\quad 3x+y-z=4$$

Solution First notice that the normals $\mathbf{N}_{1} = \mathbf{i}\, -\, 2\mathbf{j}\, +\, \mathbf{k}$ and $\mathbf{N}_{2} = 3\mathbf{i}\, +\, \mathbf{j}\, -\, \mathbf{k}$ of the two planes are not parallel. To find an equation of the line of intersection, we need a point on the line and the direction of the line. Since the line of intersection is perpendicular to both $\mathbf{N}_{1}$ and $\mathbf{N}_{2}$, it is parallel to the vector $\mathbf{D}= \mathbf{N} _{1}\times \mathbf{N}_{2}$. $$\mathbf{D=N}_{1}\times \mathbf{N}_{2}= \left|\begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 1 & -2 & 1 \\[3pt] 3 & 1 & -1 \end{array}\right| =\mathbf{i}+4\mathbf{j}+7\mathbf{k}$$

The vector $\mathbf{D}$ gives the direction of the line. We can find a point on the line by locating any point common to both planes. For example, if $z=0$, then $x-2y=-1$ and $3x+y=4$. Solving these equations simultaneously, we find $x=1$ and $y=1$. So, the point $(1,1,0)$ is on the line, and symmetric equations of the line of intersection are $$\dfrac{x-1}{1}=\dfrac{y-1}{4}=\dfrac{z}{7}$$