Given system of linear equation is

$$ \begin{gathered} 2 x-y+3 z=5 \\\\ 3 x+2 y-z=7 \\\\ 4 x+5 y+\alpha z=\beta \\\\ \text { Now, } \Delta=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{array}\right|=7(\alpha+5) \end{gathered} $$

So, this system of equation has unique solution, if $\alpha \neq-5$

and $\Delta_1=\left|\begin{array}{ccc}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha\end{array}\right|=17 \alpha-5 \beta+30$

and $\Delta_2=\left|\begin{array}{ccc}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha\end{array}\right|=-11 \beta+\alpha+104$

and $\Delta_3=\left|\begin{array}{ccc}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{array}\right|=7(\beta-9)$

For infinitely many solutions,

$$ \Delta=\Delta_1=\Delta_2=\Delta_3=0 $$

$$ \begin{aligned} & \text { So, } \Delta=0 \\\\ & \Rightarrow 7(\alpha+5)=0 \\\\ & \Rightarrow \alpha=-5 \end{aligned} $$

$$ \begin{aligned} & \Delta_3 =0 \\\\ & \Rightarrow 7(\beta-9) =0 \\\\ & \Rightarrow \beta =9 \end{aligned} $$

If $\alpha=-5$ and $\beta=8$, then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$.

Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.