Two systems 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

The planes are   $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1$and $\frac{\mathrm{x}}{{\mathrm{a}}^{\mathrm{\prime }}}+\frac{\mathrm{y}}{{\mathrm{b}}^{\mathrm{\prime }}}+\frac{\mathrm{z}}{{\mathrm{c}}^{\mathrm{\prime }}}=1$

Since the perpendicular distance of the origin on the planes is same, therefore
$\begin{array}{r}\left|\frac{-1}{\sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}}}\right|=\left|\frac{-1}{\sqrt{\frac{1}{{\mathrm{a}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{b}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{c}}^{\mathrm{\prime }2}}}}\right|\\ \mathrm{or} \frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}-\frac{1}{{\mathrm{a}}^{\mathrm{\prime }2}}-\frac{1}{{\mathrm{b}}^{\mathrm{\prime }2}}-\frac{1}{{\mathrm{c}}^{\mathrm{\prime }2}}=0\end{array}$