The vector equation of a plane which is at a distance of $5 \text{ units}$ from the origin and normal to the vector $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$ is:

If the foot of the perpendicular drawn from the point $(0, 0, 0)$ to the plane is $(4, -2, -5)$,then the equation of the plane is $......$

$A$ 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:

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z = 29$ are

If a plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,then the distance of that plane from the point $(1, 2, 2)$ is

Find the equation of the plane passing through the points $A(2, 2, -1)$,$B(3, 4, 2)$,and $C(7, 0, 6)$.