Let the lines  $y+2x=\sqrt{11}+7\sqrt{7}and2y+x=2\sqrt{11}+6\sqrt{7}$  be normal to a circle  $C:{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ . If the line  $\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$ is tangent to the circle C, then the value of  ${\left(5h-8k\right)}^2+5r^2$ is equal to _______.

Normal lines to the circle are  $y+2x=\sqrt{11}+7\sqrt{7},2y+x=22\sqrt{11}+6\sqrt{7}$ 
Center of th circle is point of intersection of normals  $\left(\frac{8\sqrt{7}}{3},\sqrt{11}+\frac{5\sqrt{7}}{3}\right)$

Given equation of tangent is  $\sqrt{11y}-3x=\frac{2\sqrt{77}}{3}+11$  
Distance of tangent from the centre is shown below.  $r=\left|\frac{8\sqrt{7}-11-\frac{5}{3}\sqrt{77}+\frac{5\sqrt{77}}{3}+11}{\sqrt{20}}\right|$

$r=\left|4\sqrt[ ]{\frac{7}{5}}\right|=4\sqrt[ ]{\frac{7}{5}}units.$

Now, ${\left(5h-8k\right)}^2+5r^2=816$