## Calculus 8th Edition

$x+y+z=4$
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, $x+y+z=4$