Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is

The shortest distance between the lines

$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is

If the plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle of $${\pi \over 2}$$, then the plane after the rotation passes through the point :

If the lines $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($$\alpha$$, 0, 0) is :