## Calculus 8th Edition

$3x-8y-z=-38$
The plane passes through the points $(1,5,1)$ and is perpendicular to the planes $2x+y-2z=2$ and $x+3z=4$. Normal vector is: $\lt 3,-8,-1\gt$ The general form of the equation of the plane is: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ or, $ax+by+cz=ax_0+by_0+cz_0$ Plugging in the values ,we get $3x-8y-z=3-40-1$ After simplification, we get $3x-8y-z=-38$