If the straight lines x−12=y+1k=z2 and x+15=y+12=zk are coplanar, then the plane (s) containing these two lines is (are)

For given lines to be coplanar, we should have 2002k252k=0⇒k=±2For k = 2, obviously the plane y+1=z is common in both lines.For k = -2, the plane is given by x−1y+1z2−2252−2=0⇒y+z+1=0