Suppose, the line $${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$$ lies on the plane $$x + 3y - 2z + \beta = 0$$. Then $$(\alpha + \beta )$$ is equal to _______.

The square of the distance of the point of intersection

of the line $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$$ and the plane $$2x - y + z = 6$$ from the point ($$-$$1, $$-$$1, 2) is __________.

Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________.

Let Q be the foot of the perpendicular from the point P(7, $$-$$2, 13) on the plane containing the lines $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$$. Then (PQ)², is equal to ___________.