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If the distance between the plane `Ax-2y+z=d.` and the plane containing the lies `(x-1)/2=(y-2)/3=(z-3)/4 and (x-2)/3=(4-3)/4=(z-4)/5` is `sqrt6,` then `|d|` is

A

`3`

B

4

C

6

D

1

Text Solution

Verified by Experts

The 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`
This plane is at a distance of `sqrt(6)` units from the plane `Ax-2y+z=d`.
`:.A=1` and `(|d-0|)/(sqrt(1+4+1))=sqrt(6)`
`implies|d|=6`
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