Answer is (b) y + z = –1
Since given lines are coplanar, so we get
k2 – 4 = 0
k = ±2
The equation of the plane containing these two lines is
(x – 1)(k2 – 4) – (y + 1)(2k – 10) + z(4 – 5k) = 0
–(y + 1)(2k – 10) + z(4 – 5k) = 0
–(y + 1)(±4 – 10) + (4 – ±10) = 0
Taking positive sign, we get
(y + 1) – z = 0
y – z = –1
Taking negative sign, we get
– (y + 1)(– 14) + 14z = 0
(y + 1) + z = 0
y + z = – 1