$$\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}$$

For infinite solutions $$\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$$ and

$$\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}$$

$$\alpha \beta-3 \alpha+4 \beta=17$$ ..... (1)

$$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}$$

$$13 \beta=54, \beta=\frac{54}{13}$$ put in (1)

$$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$$

$$54 \alpha-39 \alpha+216=221$$

$$15 \alpha=5 \quad \alpha=\frac{1}{3}$$

Now, $$12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$$

$$=4+54=58$$