Find the equation of the plane bisecting the angle between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$.

For what value of $k$ do the planes $kx + 4y + z = 0$,$4x + ky + 2z = 0$,and $2x + 2y + z = 0$ intersect in a single line?

If the foot of the perpendicular from $(0,0,0)$ to the plane is $(1,2,2)$, then the equation of the plane is

The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\pi/4$ with the plane $2x + 3y = 7$ are

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$,$(-1, 1, -2)$ and $(3, -4, 2)$ is:

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$