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

The equation of the plane having intercepts $a, b, c$ with $x, y, z$ axes respectively is given by:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ..........$(1)$
The distance $p$ of the plane from the origin $(0, 0, 0)$ is given by the formula for the distance of a point from a plane $Ax + By + Cz + D = 0$,which is $p = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = \frac{1}{a}$,$B = \frac{1}{b}$,$C = \frac{1}{c}$,and $D = -1$.
Substituting these values:
$p = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) + \frac{1}{c}(0) - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}}$
$p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$
Squaring both sides:
$p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$
Taking the reciprocal:
$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$