If the straight lines `(x-1)/(2)=(y+1)/(k)=(z)/(2) and (x+1)/(5)=(y+1)/(2)=(z)/(k)` are coplanar, then the plane(s) containing these two lines is/are

Options

A. `y+12z=-1`
B. `y+1z=-1`
C. `y+-2z=1`
D. `y+-z=1`

Answer

B

Explanation

The the given lines to be coplanar, we must have
`|(1-(-1),-1-(-1),0),(2,k,2),(5,2,k)|=0implies|(2,0,0),(2,k,2),(5,2,k)|=0`
`impliesk^(2)-4=0impliesk=+-2`
The equations of the plane containing these lines are
`|(x+1,y+1,z),(2,k,2),(5,2,k)|=0`
`implies(k^(2)-4)(x+1)-(2k-10)(y+1)+(4-5k)z=0`
`implies(k^(2)-4)x-2(k-5)y+(4-5k)z+k^(2)-4-2k+10=0`
`implies6y-6z+6=0,14y+14z+14=0` [ Putting `k=+-2`]
`impliesy-z+1=0,y+z+1=0impliesy+-=-1`