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

If the lines $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $$\,{{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then the value of $$k$$ is