Find the equation of the plane that passes through the point (3,7,-1) and is perpendicular to the line of intersection of the planes x-y-2z+3=0 and 3x-2y+z+5=0

Answer 1

The line of intersection of the planes is in the direction of the cross-product of the two normals:

{1,-1,-2}x{3,-2,1} = {-5,-7,1}

Now we have a point and a normal vector.
The equation of the plane is thus

-5(x-3) - 7(y-7) + 1(z-1) = 0
-5x + 15 - 7y + 49 + z - 1 = 0
5x+7y-z = 63