Answer
$x+y+z=4$
Work Step by Step
Planes $x-z=1$ and $y+2z=3$
Direction vectors of the line of intersection is $ \lt 1-0,3-5, 0-(-1) \gt= \lt 1,-2,1 \gt$
Parametric equations of the line can be
$x=1+t,y=3-2t,z=0+t$
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
$3(x-0)+3(y-5)+3(z-(-1))=0$
$3x+3y+3x=12$
Hence, $x+y+z=4$
