If the foot of the perpendicular drawn from the origin to a plane is $M(-1, -2, 2)$,then the vector equation of the plane is

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

The equation of the plane,passing through the point $(-1, 2, -3)$ and parallel to the lines $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y-1}{-3} = \frac{z-2}{2}$,is

$A$ point moves such that its distances from the points $(3, 4, -2)$ and $(2, 3, -3)$ are equal. What is the locus of the point?

Given points $A(3, 2, -1)$ and $B(1, 4, 3)$,find the equation of the plane that bisects the segment $AB$ perpendicularly.

$A$ variable plane passes through the fixed point $(3, 2, 1)$ and meets $X, Y,$ and $Z$ axes at points $A, B,$ 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: