Find the distance of the point $(0, 0, 0)$ from the plane $3x - 4y + 12z = 3$. (in $/13$)

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

For which values of $a$ do the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on opposite sides of the plane $3x + 4y - 12z + 13 = 0$?

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

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