multivariable calculus – plane P1 : 9 x− 3 y+ 9 z= 8. The plane P1 is perpendicular to a plane P2 . The points A:(3,2,0) and B:(2,−2,1) lie on plane P2. Plane 2 Equation?

Consider a plane P1 : 9 x− 3 y+ 9 z= 8. The plane P1 is perpendicular to a plane P2 . The points A:(3,2,0) and B:(2,−2,1) lie on plane P2.

The scalar equation of plane P2 is:

I know that P1, the normal vector is <9,-3,9> and I calculated AB which is <-1,-4,1> .

Now this where im uncertain, I calculated the cross product of AB and the normal Vector of P1 which led me to -33x – 18y +39z , but the answer according to the test was 18x +39y -63z.Where am i going wrong?