Three Dimensional Geometry Part 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)