| = 0 ⇒ k2 = 4, ⇒ k = ± 2 For k = 2, obviously the plane y + 1 = z is common in both lines. For k = – 2, the family of plane containing first line is x + y + λ (x – z – 1) = 0 Point (– 1, – 1, 0) must satisfy it - 2 + λ (- 2) = 0 ⇒ λ = - 1 ⇒ y + z + 1 = 0