Given equation of sides are
$7 x-y+3 =0 \,\,\,\,\,\,\dots(i)$
$x+y-3 =0 \,\,\,\,\,\,\dots(ii)$
Equation of angle bisectors of Eqs. (i) and (ii) are
$\frac{7 x-y+3}{\sqrt{49+1}}=\pm \frac{x+y-3}{\sqrt{1+1}}$
$\frac{7 x-y+3}{\sqrt{50}}=\pm \frac{x+y-3}{\sqrt{2}} $
$\frac{7 x-y+3}{5}=\pm \frac{x+y-3}{1} $
$\therefore \, \frac{7 x-y+3}{5}=(x+y-3)$
or $\frac{7 x-y+3}{5}=-(x+y-3)$
or $7 x-y+3=5 x+5 y-15$
$7 x-y+3=-5 x-5 y+15$
$\Rightarrow \,2 x-6 y+12=0$
or $12 x+4 y-12=0$
$\Rightarrow \, x-3 y+6=0$ or $3 x+y-3=0$
Slope $=\frac{1}{3}$, Slope $=-3$ (integer)
Here, third side is parallel to one angle bisector. $\therefore $ Required slope of third side is $-3$.
