If two planes intersect,then the shortest distance between the planes is

If the given planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are mutually perpendicular,then:

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$,and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is:

The distance of a point $(1, 2, -1)$ from the plane $x - 2y + 4z + 10 = 0$ is

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $2x + 3y - z = 5$.