JEE Mains 25-July-2022 Shift 2 Solved Paper

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=|‌

2a

√2

|
as given, |‌

2a

√2

|=3√2⇒a=±3
Hence, Q≡(±3,±3,2)
Distance 7Q=√21⇒(PQ)2=21