The product of the perpendicular from origin to the pair of lines $5{\mathrm{x}}^2+12\mathrm{xy}+6{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}-7=0$is 

The product of the perpendiculars drawn from the point $\mathrm{P}\left(0,0\right)$to the pair of lines represented by the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \mathrm{is} \left|\frac{\mathrm{c}}{\sqrt{{\left(\mathrm{a}-\mathrm{b}\right)}^2+4{\mathrm{h}}^2}}\right|$
For the equation $5{\mathrm{x}}^2+12\mathrm{xy}+6{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}-7=0$the product of the perpendiculars from the point $\left(0,0\right)$ is d
$\begin{array}{l}\mathrm{d}=\left|\frac{-7}{\sqrt{{\left(5-6\right)}^2+{\left(12\right)}^2}}\right|\\ =\left|\frac{7}{\sqrt{145}}\right|\end{array}$
Therefore, the required distance is $\boxed{\frac{7}{\sqrt{145}}}$