A plane which is perpendicular to two planes 2x-2y+ z=0 and x- y+2z=4 passes through (1,-2,1).The square of distance of the plane from the point (1,2,2) is

The equation of the plane passing through the point (1, - 2, 1) and perpendicular to the planes 2x - 2y + z = 0 and x - y + 2z= 4 is given byx−1y+2z−12−211−12=0or    x+y+1=0Its distance from the point (1,2,2) is  1+2+12=22