The plane through the intersection of the planes x+y+z=1 and 2x+3y-z+4=0 and parallel to y-axis also passes through the point. (3,2,1)

Equation of plane (x+y+z-1) + λ(2x+3y-z+4)=0 ⇒ (1+2λ)x+(1+3λ)y+(1-λ)z-1+4λ=0
dr's of normal of the plane are 1+2λ, 1+3λ, 1-λ
Since plane is parallel to y-axis, 1+3λ=0 ⇒ λ=-1/3
So the equation of plane is x+4z-7=0
Point (3,2,1) satisfies this equation
Hence Answer is (3,2,1).