Three Dimensional Geometry

For given lines to be coplanar, we get
|

2	k	2
5	2	k
2	0	0

| = 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