$$ \begin{aligned} & 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \end{aligned} $$
has infinitely many solutions, then $13 \alpha \beta$ is equal to :
Let $$A$$ be a $$3 \times 3$$ real matrix such that
$$A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \text {. }$$
Then, the system $$(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$$ has :
If the system of linear equations
$$\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned}$$
has infinitely many solutions, then $$12 \alpha+13 \beta$$ is equal to
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