Two system of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and a', b', c' from the origin, then

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a'^2}-\frac{1}{b'^2}-\frac{1}{c'^2}=0$

The equation of the plane with reference to the two systems of rectangular axes are

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

and, $\frac{X}{a'}+\frac{Y}{b'}+\frac{Z}{c'}= 1$ ............(ii)

Since the origin is same.

∴ Length of the ⊥ from (0, 0, 0) on (i)

= Length of the ⊥ from (0, 0, 0) on (ii)

$⇒ \begin{vmatrix} \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\end{vmatrix}=\begin{vmatrix} \frac{1}{\sqrt{\frac{1}{a'^2}+\frac{1}{b'^2}+\frac{1}{c'^2}}}\end{vmatrix}$

$⇒\frac{1}{a^2}+ \frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a'^2}-\frac{1}{b'^2}-\frac{1}{c'^2}=0$