$A$ plane bisects the line segment joining the points $(1, 2, 3)$ and $(-3, 4, 5)$ at right angles. Then this plane also passes through the point

The equation of a plane parallel to the plane $x - 2y + 2z - 5 = 0$ and at a unit distance from the origin is:

The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

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:

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is:

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: