$L _1: y = x +2, L _2: 4 y =3 x +6, L _3: 3 y =4 x +1$
Bisector of lines $L _2 \& L _3$
$ \frac{4 x-3 y+1}{5}=\pm\left(\frac{3 x-4 y+6}{5}\right)$
$(+) 4 x-3 y+1=3 x-4 y+6$
$ x+y=5$
Centre lies on Bisector of $4 x-3 y+1=0 \&$
(0) $3 x-4 y+6=0$
$\Rightarrow h + k =5$