A plane P contains the line $$x + 2y + 3z + 1 = 0 = x - y - z - 6$$, and is perpendicular to the plane $$ - 2x + y + z + 8 = 0$$. Then which of the following points lies on P?

For real numbers $$\alpha$$ and $$\beta$$ $$\ne$$ 0, if the point of intersection of the straight lines

$${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$$ and $${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$$, lies on the plane x + 2y $$-$$ z = 8, then $$\alpha$$ $$-$$ $$\beta$$ is equal to :

Let the plane passing through the point ($$-$$1, 0, $$-$$2) and perpendicular to each of the planes 2x + y $$-$$ z = 2 and x $$-$$ y $$-$$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :

Let the foot of perpendicular from a point P(1, 2, $$-$$1) to the straight line $$L:{x \over 1} = {y \over 0} = {z \over { - 1}}$$ be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If $$\alpha$$ is the acute angle between the lines PN and PQ, then cos$$\alpha$$ is equal to ________________.