The distance of the point $(2, -1, 0)$ from the plane $2x + y + 2z + 8 = 0$ is

The distance between the two planes $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is

If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x+py+qz+r=0$,then $\sqrt{p^2+q^2+r^2}=$

$A$ 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:

Find the distance of a point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$. (in $/7$)

$A$ point $P(a, a, a)$ lies on a line passing through the origin and making equal angles with the axes,where $a$ is a constant. $A$ plane passes through $P$ and is perpendicular to $OP$. If this plane makes intercepts on the axes,what is the sum of the reciprocals of these intercepts?