The distance of the point $(2, -1, 0)$ from the plane $2x + y + 2z + 8 = 0$ is

If $(3,4,-7)$ is the foot of the perpendicular drawn from the point $(-2,3,6)$ to the plane $\pi$,then the sum of the intercepts made by the plane $\pi$ on the $X$ and $Y$-axes is

$A$ variable plane is at a distance $k$ from the origin and meets the coordinate axes at $A, B, C$. The locus of the centroid of $\Delta ABC$ is . . . . . .

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

Find the equation of the plane passing through the intersection of the planes $3x - y + 2z - 4 = 0$ and $x + y + z - 2 = 0$ and the point $(2, 2, 1)$.

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$