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

<p>${L_1}:y + 2x = \sqrt {11} + 7\sqrt 7$</p> <p>${L_2}:2y + x = 2\sqrt {11} + 6\sqrt 7$</p> <p>Point of intersection of these two lines is centre of circle i.e. $\left( {{8 \over 3}\sqrt 7 ,\sqrt {11} + {5 \over 3}\sqrt 7 } \right)$</p> <p>${ \bot ^r}$ from centre to line $3x - \sqrt {11} y + \left( {{{5\sqrt {77} } \over 3} + 11} \right) = 0$ is radius of circle</p> <p>$$ \Rightarrow r = \left| {{{8\sqrt 7 - 11 - {5 \over 3}\sqrt {77} + {{5\sqrt {77} } \over 3} + 11} \over {\sqrt {20} }}} \right|$$</p> <p>$= \left| {\root 4 \of {{7 \over 5}} } \right| = \root 4 \of {{7 \over 5}}$ units</p> <p>So ${(5h - 8K)^2} + 5{r^2}$</p> <p>$$ = {\left( {{{40} \over 3}\sqrt 7 - 8\sqrt {11} - {{40} \over 3}\sqrt 7 } \right)^2} + 5.\,16.\,{7 \over 5}$$</p> <p>$= 64 \times 11 + 112 = 816$.</p>