If the distance between the plane `Ax-2y+z=d and ` the plane containing the lines `(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-3)/(4)=(z-4)/(5)` is `sqrt(6)`, then |d| is equal to….

Options

A. 3
B. 4
C. 5
D. 6

Answer

D

Explanation

Equation of the plane containing the given lines is. `|{:(x-1,y-2,z-3,),(2,3,4,),(3,4,5,):}|=0impliesx-2y+z=0`
since this plane must be parallel to the plane `Ax-2y+z=d`, we get `A=1` As distance between the planes is `sqrt(6)`.
we get `(|d|)/(sqrt(1+4+1))sqrt(6)implies|d|=6`