Answer
$3x-8y-z=-38$
Work Step by Step
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
$3x-8y-z=-38$
