In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

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

If the foot of the perpendicular drawn from the origin to the plane is $(3, 2, 1)$,then the equation of the plane is

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$,then the equation of the plane is

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are