Find the equation of the plane in xyz-space through the point P = (4, 5, 3) and perpendicular to the vector n = (2, 1, 1). z = Solution The equation of a plane passing through P with normal vector n is n · (x - P) = 0. So, the plane in question is given by the cross product. <2, 1, 1> · = 0 ==> 2(x - 4) + 1(y - 5) - 1(z - 3) = 0..