The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is

If $l, m, n$ are the direction cosines of a normal to the plane passing through the points $(0, 1, 2)$, $(3, 0, 2)$, and $(4, 5, 0)$, then $|l| + |m| + |n| = $

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are

$A$ bisector of the angle between the normals of the planes $4x + 3y = 5$ and $x + 2y + 2z = 4$ is along the vector

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

$A$ plane is at a unit distance from the origin. It cuts the coordinate axes at $P, Q,$ and $R$ respectively. If the locus of the centroid of the $\Delta PQR$ is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,then $k =$