The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :

The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :

Consider the lines $$L_1$$ and $$L_2$$ given by

$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.

A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is

If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha,\beta,\gamma$$), then $$\alpha+\beta+\gamma$$ is equal to :