Find equation of plane of the given question

Question

Find equation of plane through $P(-2,1,7)$ that is perpendicular to the plane $4x-2y+2z=-1$ and $3x+3y-6z=5$

Answer 1

$N_1= \langle4,-2,2\rangle$ and $N_2= \langle3,3,-6\rangle$ are normal vectors to the given planes.

The common perpendicular $N=N_1\times N_2$ is the normal vector to the new plane.

$N= \langle6,30,18\rangle$, thus the equation of the desired plane is $$6x+30y+18Z=d$$

Where $d=144$ is found from the given point.

Thus we have $$ 6x +30y+18Z=144$$ which simplifies to $$ x +5y+3Z=48$$

Answer 2

Hint:

The normal vector of your new plane, must be perpendicular to both of the normal vectors of the previous two planes.