The distance between the circumcentre and the orthocentre of the triangle whose vertices are $\left(0,0\right),\left(6,8\right),$ and $\left(-4,3\right)$ is L. Then the value of $\frac{2}{\sqrt{5}}\mathrm{L}$ is __

Given vertices of triangle are $\mathrm{O}\left(0,0\right),\mathrm{B}\left(6,8\right)$ and $\mathrm{C}\left(-4,3\right)$. 

Slope of $\mathrm{OB}=\frac{8}{6}=\frac{4}{3}$

Slope of $\mathrm{OC}=\frac{3}{4}$

$\therefore \angle \mathrm{BOC}=\frac{\mathrm{\pi }}{2}$

$\mathrm{\Delta BOC}$ is right -angled at O.

Circumcenter=Midpoint of hypotenuse $\mathrm{BC}=\left(1,\frac{11}{2}\right)$

Required distance  $=\sqrt{\left(1+\frac{121}{4}\right)}=\frac{5\sqrt{5}}{2}\mathrm{units}$