Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$                                                                   $\ldots(1)$

The distance (p) of the plane from the origin is given by,

$p=\left|\frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}+\left(\frac{1}{c}\right)^{2}}}\right|$

$\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}$

$\Rightarrow p^{2}=\frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$

$\Rightarrow \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$