[Solved] 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 \\mid d \\mid is - - JEE (Advanced) 2010 Paper-1