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

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