A chord AB which is bisected at (1, 1) is drawn to the hyperbola  $7x^2+8xy-y^2-4=0$ with centre C, which intersects the asymptotes in E and F. If the equation of circumcircle of  $\Delta \mathcal{l}e$ CEF is  $x^2+y^2-ax-by+c=0$ then the value of  $\frac{11a+21b+7c}{56}$ is equal to ____________    

Equation of chord AB is  $S_1=S_{11}$ whose mid point is (1, 1)
 $\Rightarrow 7x\left(1\right)+4(x\left(1\right)+\left(y\right)\left(1\right))-y\left(1\right)-4=7+8-1-4$
 $\Rightarrow 11x+3y=14$
 
Equation of circumcircle CEF is  $x^2+y^2-ax-by+c=0$
Homogenising circumcircle with the line EF the asymptotes are obtained.
 $x^2+y^2-ax\left(1\right)-by\left(1\right)+c{\left(1\right)}^2=0$
 $x^2+y^2-ax\left(\frac{11x+3y}{14}\right)-by\left(\frac{11x+3y}{14}\right)+c{\left(\frac{11x+3y}{14}\right)}^2=0$
 $\Rightarrow x^2[1-\frac{11a}{14}+\frac{121c}{196}]+y^2[1-\frac{3b}{14}+\frac{9c}{196}]+xy[\frac{-3a}{14}-\frac{11b}{14}+\frac{66c}{196}]=0$
Equation of asymptotes is  $7x^2+8xy-y^2=0$
 $\frac{1-\frac{11a}{14}+\frac{121c}{196}}{7}=\frac{1-\frac{3b}{14}+\frac{9c}{196}}{-1}=\frac{\frac{-3a}{14}-\frac{11b}{14}+\frac{66c}{196}}{8}$
a = 0, a = $\frac{112}{23}$ , b = $\frac{64}{23}$ 
 $\frac{1-\frac{11a}{14}}{7}=\frac{1-\frac{3b}{14}}{-1}$
 $\Rightarrow 11a+21b=112$
 $\Rightarrow \frac{11a+21b+7c}{56}=\frac{112}{56}=2$