Two systems of rectangular axes have the same origin If a plane cuts them at the distance a, b, c and a’, b’, c’ respectively from the origin, then
$\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}=\mathrm{k}\left(\frac{1}{{\mathrm{a}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{b}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{c}}^{\mathrm{\prime }2}}\right),\text{ where }\mathrm{k}=$

Let a, b, c be the intercepts when Ox, Oy, Oz are taken as axes: 
Then the equation of the plane is $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1$
Also let a’, b’, c’, be the intercepts when OX, OY,OZ are taken as axes ;
Then in this case equation of the same plane is $\frac{\mathrm{X}}{{\mathrm{a}}^{\mathrm{\prime }}}+\frac{\mathrm{Y}}{{\mathrm{b}}^{\mathrm{\prime }}}+\frac{\mathrm{Z}}{{\mathrm{c}}^{\mathrm{\prime }}}=1$
Now (1) and (2) are equations of the same plane and in both the cases the origin is
same. Hence length of the perpendicular drawn form the origin to the plane in both
the case must be the same
i.e., $\frac{1}{\sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}}}=\frac{1}{\sqrt{\frac{1}{{\mathrm{a}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{b}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{c}}^{\mathrm{\prime }2}}}}$
or $\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}=\frac{1}{{\mathrm{a}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{b}}^{\mathrm{\prime }2}}+\frac{1}{{\mathrm{c}}^{\mathrm{\prime }2}}\therefore \mathrm{k}=1$