Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :

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