If the line ${\mathrm{l}}_1:3\mathrm{y}-2\mathrm{x}=3$  is the angular bisector of the lines  ${\mathrm{l}}_2:\mathrm{x}-\mathrm{y}+1=0$ and  ${\mathrm{l}}_3:\mathrm{\alpha x}+\mathrm{\beta y}+17=0$ , then ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-\mathrm{\alpha }-\mathrm{\beta }$ is equal to_______.

${\mathrm{l}}_1:3\mathrm{y}-2\mathrm{x}=3$
${\mathrm{l}}_2:\mathrm{x}-\mathrm{y}+1=0$

Solving , ${\mathrm{l}}_1,{\mathrm{l}}_2$, A = (0,1) lies on ${\mathrm{l}}_3:\mathrm{\alpha x}+\mathrm{\beta y}+17=0$

$\Rightarrow \mathrm{\beta }=-17$
We have (-1,0) lies on ${\mathrm{l}}_2$
Let (h,k) be the image of (-1,0) is ${\mathrm{l}}_1$

$\frac{\mathrm{h}+1}{2}=\frac{\mathrm{k}}{-3}=\frac{-2\left(1\right)}{13}\Rightarrow (\mathrm{h},\mathrm{k})=(\frac{-17}{13},\frac{6}{13})$

(h,k) lies on ${\mathrm{l}}_3\Rightarrow \mathrm{\alpha }\left(\frac{-17}{13}\right)-17\left(\frac{6}{13}\right)+17=0$

$\Rightarrow \mathrm{\alpha }=7$

$\therefore {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-\mathrm{\alpha }-\mathrm{\beta }=348$