$=1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha)$

$=\alpha \beta+3+4 \beta-18-2-3 \alpha$

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

$D_3=0$

 $D=0$

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

$D_1=\left|\begin{array}{ccc}-4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta\end{array}\right|=0$

$D_2=\left|\begin{array}{ccc}1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta\end{array}\right|=0$

$\Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0$

$13 \beta-9-36-9=0$

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

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

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

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

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

$=4+54=58$