Consider three planes $$${P_1}:x - y + z = 1$$$ $$${P_2}:x + y - z = 1$$$ $$${P_3}:x - 3y + 3z = 2$$$

Let $${L_1},$$ $${L_2},$$ $${L_3}$$ be the lines of intersection of the planes $${P_2}$$ and $${P_3},$$ $${P_3}$$ and $${P_1},$$ $${P_1}$$ and $${P_2},$$ respectively.

STATEMENT - 1Z: At least two of the lines $${L_1},$$ $${L_2}$$ and $${L_3}$$ are non-parallel and

STATEMENT - 2: The three planes doe not have a common point.

Consider the planes $$3x-6y-2z=15$$ and $$2x+y-2z=5.$$

STATEMENT-1: The parametric equations of the line of intersection of the given planes are $$x=3+14t,y=1+2t,z=15t.$$ because

STATEMENT-2: The vector $${14\widehat i + 2\widehat j + 15\widehat k}$$ is parallel to the line of intersection of given planes.

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$(1, -2, 1).$$ The distance of the plane from the point $$(1, 2, 2)$$ is

A variable plane at a distance of the one unit from the origin cuts the coordinates axes at $$A,$$ $$B$$ and $$C.$$ If the centroid $$D$$ $$(x, y, z)$$ of triangle $$ABC$$ satisfies the relation $${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = k,$$ then the value $$k$$ is