Find the distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$. (in $/3$)

The distance from a point $(1, 1, 1)$ to a variable plane $\pi$ is $12$ units and the points of intersections of the plane $\pi$ and $X, Y, Z$-axes are $A, B, C$ respectively. If the point of intersection of the planes through the points $A, B, C$ and parallel to the coordinate planes is $P$,then the equation of the locus of $P$ is

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$. Hence,the equation of the plane is

Equations of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of $1$ unit from the point $(1, 2, 3)$ are $.....$

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

$A$ variable plane passes through a fixed point $(3, 2, 1)$ and meets the $x, y,$ and $z$ axes at $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $yz$-plane through $A$,a second plane is drawn parallel to the $zx$-plane through $B$,and a third plane is drawn parallel to the $xy$-plane through $C$. Then,the locus of the point of intersection of these three planes is: