3 points lie in a plane; P1=i+2j+3k, P2=-3i+5j+2k, P3=i+2j+k. Find the Cartesian equation of the plane

Take 2 vectors on the plane, originating from the same point, that aren't parallel: P1-->P2 = P2-P1 = -4i+3j-kP1-->P3 = P3-P1 = 0i+0k-2kOne can then find the cross product of these two vectors to determine a vector that is always perpendicular to the plane. (P1-->P2)x(P1-->P3) =i(3*-2 - -10) - j(-4-2 - -10) + k(-40 - 30) = -6i -8j +0k The general form of the plane can then be given by n⋅r=n⋅a where n is the normal to the plane, r is any point, and a is any point on the plane. Hence the general form of the plane can be given by(-6i-8j+0k)⋅r=(-6i-8j+0k)⋅(1i+2j+3k)=-61+-82+03=22This equation can be converted to cartesian form, so that-6x-8y+0z=22