If the distance between the planes `Ax-2y+z=d` and the plane containing the lines `(x - 1) / 2=(y - 2) /3=(z - 3)/4` and

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Q 1124191051.     If the distance between the planes `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 `sqrt6`, then `|d|`
JEE 2010
A

`6`

B

`7`

C

`8`

D

`9`

HINT

The plane containing the two lines be `px+qy+rz=k` So the normal to plane is normal to the lines lying on the plane.

Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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