Find an equation of the plane through the points (3,4,-2) and perpendicular to the vector 2x-3y-z
normal vector = 2i -3 j -1 k
so 2(x-xo) -3(y-yo) -1(z-zo) = 0
2 x -3 y -1z = 2(3)-3(4)-1(-2) = 6 -12 + 2 = -4
so 2(x-xo) -3(y-yo) -1(z-zo) = 0
2 x -3 y -1z = 2(3)-3(4)-1(-2) = 6 -12 + 2 = -4