y=x2+px−3 Let P(α,0),Q(β,0),R(0,−3) Circle with centre (−1,−1) is (x+1)2+(y+1)2=r2 Passes through (0,−3) $\begin{aligned}
& \left.1^2+(-2)^2=r^2\right] \
& \mathrm{r}^2=5
\end{aligned}(x+1)^2+(y+1)^2=5Put\mathrm{y}=0\begin{aligned}
& (x+1)^2=5-1 \
& (x+1)^2=4 \
& x+1= \pm 2 \
& x=1 \text { or } x=-3 \
& \therefore P(1,0) \text { and } Q(-3,0)
\end{aligned}Areaof\triangle \mathrm{PQR}=\frac{1}{2}\left|\begin{array}{ccc}1 & 0 & 1 \ -3 & 0 & 1 \ 0 & -3 & 1\end{array}\right|=6$