Let $\text{P}=\{\text{z}\in ℂ:|\text{z}+2-3\text{i}|\le 1\}$ and $Q=\{z\in ℂ:z(1+i)+\overline{z}(1-i)\le -8\}.$ Let in $\text{P}\cap \text{Q},|\text{z}-3+2\text{i}|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If ${\left|z_1\right|}^2+2|z_2{|}^2=\alpha +\beta \sqrt{2},$ where $\alpha ,\beta$ are integers, then $\frac{\alpha +\beta }{4}$ equals____.

 
Clearly for the shaded region ${\text{z}}_1$ is the intersection of the circle and the line passing through $\text{P}\left({\text{L}}_1\right)$ and ${\text{z}}_2$ is intersection of line ${\text{L}}_1{\text{\&L}}_2$  
Circle : ${(x+2)}^2+{(y-3)}^2=1$ 
${\text{L}}_1:\text{x}+\text{y}-1=0, {\text{L}}_2:\text{x}-\text{y}+4=0$ 
On solving circle & ${\text{L}}_1$ we get ${\text{z}}_1:(-2-\frac{1}{\sqrt{2}},3+\frac{1}{\sqrt{2}})$
On solving ${\text{L}}_1$ and ${\text{z}}_2$ is intersection of line ${\text{L}}_1{\text{\&L}}_2$  
we get $z_2:(\frac{-3}{2},\frac{5}{2})$ 
${\left|z_1\right|}^2+2{\left|z_2\right|}^2=14+5\sqrt{2}+17=31+5\sqrt{2}$ 
So $\alpha =31, \beta =5$  
$\alpha +\beta =36$