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### Question Asked by a Student from EXXAMM.com Team

** 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 **

#### 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.