Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$ respectively from the origin,prove that $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}$.

(N/A) Let the two systems of rectangular axes be $S_1$ and $S_2$ with the same origin $O$.
The equation of the plane in the first system $S_1$ is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The equation of the plane in the second system $S_2$ is given by $\frac{x}{a^{\prime}} + \frac{y}{b^{\prime}} + \frac{z}{c^{\prime}} = 1$.
The perpendicular distance $p$ from the origin $(0, 0, 0)$ to a plane $\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$ is given by $p = \frac{|-1|}{\sqrt{\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2}}} = \frac{1}{\sqrt{\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2}}}$.
Since the plane is the same,the perpendicular distance $p$ from the origin must be the same for both systems.
Therefore,$\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = \frac{1}{\sqrt{\frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}}}$.
Squaring both sides,we get $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} = \frac{1}{\frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}}$.
Taking the reciprocal of both sides,we obtain $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}$.
Hence proved.