First plane, P1=2x−2y+z=0, normal vector ≡n1=(2,−2,1) Second plane, P2≡x−y+2z=4, normal vector ≡n2=(1,−1,2) Plane perpendicular to P1 and P2 will have normal vector n3 Where n3=(n1×n2) Hence, n3=(−3,−3,0) Equation of plane E through P(1,−1,1) and n3 as normal vector (x−1,y+1,z−1)⋅(−3,−3,0)=0 ⇒x+y=0≡E Distance of PQ(a,a,2) from E=|‌