The equation $xy = 0$ in three-dimensional space represents

The distance of the plane $6x - 3y + 2z - 14 = 0$ from the origin is

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

$A$ plane passing through the points $A(2, 3, 5)$ and $B(-3, -5, -7)$ is perpendicular to the plane $x - y + z = 1$. Which of the following points lies on this plane?

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i}-3 \hat{j}+4 \hat{k} .$ Also find its cartesian form.

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