$A$ 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:

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of $\triangle ABC$ is at the point $(1, 2, 3)$,then the equation of the plane is

The Cartesian equation of a plane which passes through the points $A(2,2,2)$ and makes equal non-zero intercepts on the coordinate axes is

Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin,then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}$.

$A$ plane $\pi$ given by $ax + by + 11z + d = 0$ is perpendicular to the planes $2x - 3y + z = 4$ and $3x + y - z = 5$. The perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive,then $d =$

The equation of the plane passing through the midpoint of the line segment joining the points $A(4, 5, -10)$ and $B(-1, 2, 1)$ and perpendicular to $AB$ is: