Answer
$x=-5+5t; y=3-t; z=-3t$
Work Step by Step
Since, we have $x+2y=1$ and $x-y=-8$
After solving, we get $y=3$ and $x=-8+y=-8+3=5$
The normal to the plane is: $n=\lt 5,-1,-3 \gt$
We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Then, for the point $(-5,3,0 )$, we have the parametric equations:
$x=-5+5t; y=3-t; z=0-3t=-3t$
Hence, $x=-5+5t; y=3-t; z=-3t$
