If the line, $${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, $${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$$ is :

The coordinates of the foot of the perpendicular from the point (1, $$-$$2, 1) on the plane containing the lines, $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7},$$ is :

The line of intersection of the planes $$\overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\,$$ and
$$\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2,$$ is :

The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal perpendicular to both the lines

$${{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}$$

and

$${{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}$$ is :