Since 3(1) + 2(– 2) + (–1) (–1) = 3 – 4 + 1 = 0 ∴ Given line is ⊥ to the normal to the plane i.e., given line is parallel to the given plane. Also, (1, –1, 3) lies on the plane x – 2y – z = 0 if 1 – 2 (–1) – 3 = 0 i.e., 1 + 2 – 3 = 0 which is true ∴ L lies in plane π.