Answer
$x+5y+3z = -6$
Work Step by Step
Given planes:
\begin{align*}
2x-y+z &= 1 \\
x+y-2z &= 3
\end{align*}
Normal vectors to the given planes:
\begin{align*}
\mathbf{u} &= \langle 2, -1, 1 \rangle \\
\mathbf{v} &= \langle 1, 1, -2 \rangle
\end{align*}
Equation of the plane parallel to the line of intersection between the given planes:
\begin{align*}
&a(x+1) + b(y-2) + c(z+5) = 0 \\
\mathbf{n} &= \mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -1 & 1 \\
1 & 1 & -2
\end{vmatrix} = \langle 1, 5, 3 \rangle \\
a &= 1, \quad b = 5, \quad c = 3 \\
&\therefore x+5y+3z = -6
\end{align*}
