Let the equation of the plane in both the systems be

$ \large\frac{x}{a}+\large\frac{y}{b}+\large\frac{z}{c}$$=1$ and $ \large\frac{X}{a'}+\large\frac{Y}{b'}+\large\frac{Z}{c'}$$=1$

We know that $\perp$ diatance of the plane $ax+by+cz+d=0$ from origin is $\bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$

It is given that the origin is same for both the system.

Hence $\perp$ distance of the plane from origin are equal for both the system.

$(ie)\Bigg| \Large\frac{-1}{\sqrt{\Large\frac{1}{a^2}+\Large\frac{1}{b^2}+\Large\frac{1}{c^2}}} \Bigg|$$ = \Bigg| \Large\frac{-1}{\sqrt{\Large\frac{1}{a'^2}+\Large\frac{1}{b'^2}+\Large\frac{1}{c'^2}}} \Bigg| $

$\Rightarrow\: \large\frac{1}{a^2}+\large\frac{1}{b^2}+\large\frac{1}{c^2}=\large\frac{1}{a'^2}+\large\frac{1}{b'^2}+\large\frac{1}{c'^2}$

Hence proved.