Find the vector and Cartesian equations of the plane that passes through the point $(1, 0, -2)$ and the normal to the plane is $\hat{i} + \hat{j} - \hat{k}$.

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

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

The distance between the parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$ 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 $|3l+2m+5n|=$

The origin lies in the acute angle between the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ if $aa' + bb' + cc' < 0$ and: