Let OA, OB, OC be coterminous edges of a cuboid. If l, m, n be the shortest distances between the sides and their respective skew body diagonals to them, respectively, then the value of $\frac{\left(\frac{1}{{\mathrm{l}}^2}+\frac{1}{{\mathrm{m}}^2}+\frac{1}{{\mathrm{n}}^2}\right)}{\left(\frac{1}{{\mathrm{OA}}^2}+\frac{1}{{\mathrm{OB}}^2}+\frac{1}{{\mathrm{OC}}^2}\right)}$ is _______

Let $\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}$ be the unit vectors along OA, OB, OC

CN, BM are diagonals skew to OA

Let 1 be the S.D. between OA and CN and let

$\overrightarrow{\mathrm{OA}}=\mathrm{x}\hat{\mathrm{i}};\overrightarrow{\mathrm{OC}}=\mathrm{z}\hat{\mathrm{k}}; \overrightarrow{\mathrm{OB}}=\mathrm{y}\hat{\mathrm{j}}\Rightarrow \overrightarrow{\mathrm{ON}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}$

Equation of OA and CN are $\overrightarrow{\mathrm{r}}=\mathrm{x}\hat{\mathrm{i}}$ and $\hat{\mathrm{r}}=\mathrm{z}\hat{\mathrm{k}}+\mathrm{p}(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}-\mathrm{z}\hat{\mathrm{k}})$ respectively.

$\therefore$ The shortest distance between those two lines.

$\mathrm{l}=\frac{\mathrm{yz}}{\sqrt{{\mathrm{y}}^2+{\mathrm{z}}^2}}\Rightarrow \frac{1}{{\mathrm{l}}^2}=\frac{1}{{\mathrm{y}}^2}+\frac{1}{{\mathrm{z}}^2}$

Similarly $\frac{1}{{\mathrm{m}}^2}=\frac{1}{{\mathrm{z}}^2}+\frac{1}{{\mathrm{x}}^2}, \frac{1}{{\mathrm{n}}^2}=\frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{y}}^2},$

$\Rightarrow \frac{1}{{\mathrm{l}}^2}+\frac{1}{{\mathrm{m}}^2}+\frac{1}{{\mathrm{n}}^2}=2\left(\frac{1}{{\mathrm{OA}}^2}+\frac{1}{{\mathrm{OB}}^2}+\frac{1}{{\mathrm{OC}}^2}\right)$