Official solution
Equation of planes be \[{x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1\]
So the distance from (0, 0, 0) to both the plane is same.
\[ \therefore \] \[\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }}} \right| = \left| {{{ - 1} \over {\sqrt {{1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}}} }}} \right|\]
\[ \Rightarrow \] \[{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {{{a'}^{2}}}} - {1 \over {{{b'}^{2}}}} - {1 \over {{{c'}^{2}}}} = 0\]