If $\alpha, \beta$ and $\gamma(\alpha<\beta<\gamma)$ are the values of $x$ such that $\left[\begin{array}{ccc}x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2 x-1\end{array}\right]$ is a singular matrix, then $2 \alpha+3 \beta+4 \gamma$ is equal to

If the set of equations $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+a z=b$ has unique solution, then