If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to

The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ?

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are :

On which of the following lines lies the point of intersection of the line,   $${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$$  and the plane, x + y + z = 2 ?