The symmetric form of the line $x+y-z=1$ and $2x-3y+z=2$ is

The given planes are  $x+y-z=1$and $2x-3y+z=2$ 

Plug in $z=0$ in equations, and then solve for $x,y$

$x+y=1,2x-3y=2$

by solving the above two equations $x=1,y=0$

The direction ratios are proportional to the vector $n_1\times n_2=\left|\begin{array}{ccc}i& j& k\\ 1& 1& -1\\ 2& -3& 1\end{array}\right|=i(-2)-j\left(3\right)+k(-5)$

Therefore, the direction ratios of the required line are proportional to $⟨2,3,5⟩$

Therefore, the equation of the line in symmetric form is $\frac{x-1}{2}=\frac{y}{3}=\frac{z}{5}$