The equation of any plane through the intersection of the planes, 3x−y+2z−4=0 and x+y+z−2=0, is 3x−y+2z−4)+λ(x+y+z−2)=0 ....(i) The plane passes through the point (2,2,1) Therefore, this point will satisfy Eq. (i). ∴(3×2−2+2×1−4)+λ(2+2+1−2)=0 ⇒(6−4)+3λ=0 ⇒2+3λ=0 ⇒ λ=
−2
3
On substituting this value of λ in Eq. (i), we get the required plane as (3x−y+2z−4)−
2
3
(x+y+z−2)=0 ⇒ 9x−3y+6z−12−2x−2y−2z+4=0 ⇒ 7x−5y+4z−8=0 This is the required equation of the plane.