Let the plane containing the line of intersection of the planes

P₁ : $$x+(\lambda+4)y+z=1$$ and

P₂ : $$2x+y+z=2$$

pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of

the point (2$$\lambda,\lambda,-\lambda$$) from the plane P₂ is :

The distance of the point (7, $$-$$3, $$-$$4) from the plane passing through the points (2, $$-$$3, 1), ($$-$$1, 1, $$-$$2) and (3, $$-$$4, 2) is :

The distance of the point ($$-1,9,-16$$) from the plane

$$2x+3y-z=5$$ measured parallel to the line

$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :

Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :