If the line $l_1:3y-2x=3$ is the angular bisector of the lines $l_2:x-y+1=0$ and $l_3:\alpha x+\beta y+17=0$, then ${\alpha }^2+{\beta }^2-\alpha -\beta$ is equal to ............

Given,

$L_1:3y-2x=3$ is angular bisector of $L_2:x-y+1=0$ and $L_3:\alpha x+\beta y+17=0$

Now finding point of intersection of $L_1 \& L_2$ we get, $\left(0, 1\right)$

And point will lie on $L_3$, so $\alpha \times 0+\beta \times 1+17=0$

$\Rightarrow \beta =-17$
Any point, say $\left(\frac{-3}{2},0\right)$ on $L_1$ should be equidistant from lines $L_2 \& L_3$

Now using the formula of distance of a point from the line we get,
$\Rightarrow \left|\frac{\frac{-3}{2}-0+1}{\sqrt{1^2+1^2}}\right|=\left|\frac{\frac{-3\alpha }{2}+0+17}{\sqrt{{\alpha }^2+(-17{)}^2}}\right|$

$\Rightarrow \left(\alpha -7\right) \left(\alpha -17\right)=0$

Now, for $\alpha =17, L_{2 }\& L_3$ coincides

So, $\alpha =7$

Now putting the value of $\alpha \& \beta$ in given expression we get,

${\alpha }^2+{\beta }^2-\alpha -\beta ={\left(-17\right)}^2+7^2-7+17$$=348$