$$\begin{aligned} &x-y+1=0\\ &\mathrm{p}=\mathrm{r}\\ &\left|\frac{\alpha-0+1}{\sqrt{2}}\right|=r \Rightarrow(\alpha+1)^2=2 r^2\quad\text{.... (1)} \end{aligned}$$

$$\begin{aligned} & \text { now }\left(\frac{-3 \alpha+0-1}{\sqrt{9+4}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=\mathrm{r}^2 \\ & \Rightarrow(3 \alpha+1)^2+4=13 \mathrm{r}^2 \ldots \ldots .(2) \\ & \text { (1) & }(2) \Rightarrow(3 \alpha+1)^2+4=13 \frac{(\alpha+1)^2}{2} \\ & \quad \Rightarrow 18 \alpha^2+12 \alpha+2+8=13 \alpha^2+26 \alpha+13 \\ & \Rightarrow 5 \alpha^2-14 \alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow \alpha=\frac{-1}{5}, 3 \end{aligned}$$

$$\begin{aligned} &\therefore \quad r=2 \sqrt{2}\\ &\text { How } \alpha \mathrm{e}=3 \text { and } 2 \alpha=4 \sqrt{2}\\ &\begin{aligned} & \alpha^2 \mathrm{e}^2=9 \Rightarrow \alpha=2 \sqrt{2} \Rightarrow \alpha^2=8 \\ & \alpha^2\left(1+\frac{\beta^2}{\alpha^2}\right)=9 \\ & \alpha^2+\beta^2=9 \\ & \therefore \beta^2=1 \\ & \therefore 2 \alpha^2+3 \beta^2=2(8)+3(1)=19 \end{aligned} \end{aligned}$$