Given:

The system of linear equations 

$x-2y+z=-4$, 

$2x+\alpha y+3z=5$ and

$3x-y+\beta z=3$ has infinity many solutions,

So, $△={△}_1={△}_2={△}_3$

Now, finding, 

${△}_2=\left|\begin{array}{ccc}1& -4& 1\\ 2& 5& 3\\ 3& 3& \beta \end{array}\right|=0$

$\Rightarrow 5\beta -9+8\beta -36-9=0$

$\Rightarrow 13\beta =54$

And, ${△}_3=\left|\begin{array}{ccc}1& -2& -4\\ 2& \alpha & 5\\ 3& -1& 3\end{array}\right|=0$

$\Rightarrow 3\alpha +5+12-30+8+12\alpha =0$

$\Rightarrow 15\alpha =5$

$\Rightarrow 12\alpha =4$

Hence, the value of $12\alpha +13\beta =4+54=58$