Since, equation of plane through intersection of planes x+y+z=1 and 2x+3y−z+4=0 is (2x+3y−z+4)+λ(x+y+z−1)=0 (2+λ)x+(3+λ)y+(−1+λ)z+(4−λ)=0 ......(1) But, the above plane is parallel to y -axis then (2+λ)×0+(3+λ)×1+(−1+λ)×0=0 ⇒λ=−3 Hence, the equation of required plane is −x−4z+7=0 ⇒x+4z−7=0 Therefore, (3,2,1) the passes through the point.