The shortest distance between the lines

$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and

$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :

A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point :

The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
$$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$$ is :