The equation of the plane which bisects the line joining $(2, 3, 4)$ and $(6, 7, 8)$ is

The equation of a 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 $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

If a plane cuts the coordinate axes at $A, B$ and $C$ respectively such that the centroid of the triangle $ABC$ is $(6, 6, 3)$, then find the equation of that plane.

Direction ratios of the normal to a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\frac{\pi}{4}$ with the plane $x + y - 3 = 0$ are:

The mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is