## Thomas' Calculus 13th Edition

$x=-5+5t; y=3-t; z=-3t$
The standard equation of a plane passing through the point $(x_1,y_1,z_1)$ can be defined as: Given: $x+2y=1$ and $x-y=-8$ Simplify the given two equations, we have $y=3$ and $x=-8+y=-8+3=5$ The equation of a vector normal to the plane is given as: $n=\lt 5,-1,-3 \gt$ Now, the parametric equations for point $(-5,3,0 )$ are: $x=-5+5t; y=3-t; z=0-3t=-3t$ Hence, $x=-5+5t; y=3-t; z=-3t$