__Ask a Question__

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

** Q 1609680518. ** If the distance between the plane `Ax-2y+z=d` and the plane containing the lines `\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}` and `\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}` is `\sqrt{6}`, then `|d|` is

**JEE 2010 ADVANCED Paper 1**