Let the acute angle bisector of the two planes x $$-$$ 2y $$-$$ 2z + 1 = 0 and 2x $$-$$ 3y $$-$$ 6z + 1 = 0 be the plane P. Then which of the following points lies on P?

The distance of line $$3y - 2z - 1 = 0 = 3x - z + 4$$ from the point (2, $$-$$1, 6) is :

The distance of the point ($$-$$1, 2, $$-$$2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x $$-$$ 2y + z = 0 is :

Let the equation of the plane, that passes through the point (1, 4, $$-$$3) and contains the line of intersection of the
planes 3x $$-$$ 2y + 4z $$-$$ 7 = 0
and x + 5y $$-$$ 2z + 9 = 0, be
$$\alpha$$x + $$\beta$$y + $$\gamma$$z + 3 = 0, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :