$\begin{aligned}
& \therefore\left(\frac{7}{3}+\beta\right) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\alpha+\gamma) \hat{\mathrm{k}}=\frac{5}{3}(\alpha \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\
& +\mathrm{B}(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+(\hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\
& \Rightarrow\left(\frac{7}{3}+\beta\right) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\alpha+\gamma) \hat{\mathrm{k}}=\left(\frac{5}{3} \alpha+1\right) \hat{\mathrm{i}} \\
& +\left(\frac{5}{3}+2 \beta+\gamma\right) \hat{\mathrm{j}}+\left(\frac{-5}{3}+\beta+3\right) \hat{\mathrm{k}}
\end{aligned}$
comparing both sides, we get
$\begin{aligned}
& \frac{7}{3}+\beta=\frac{5}{3} \alpha+1 \Rightarrow 5 \alpha-3 \beta=4 \\
& -1=\frac{5}{3}+2 \beta+\gamma \Rightarrow 2 \beta+\gamma=\frac{-8}{3} \\
& \alpha+\gamma=\frac{-5}{3}+\beta+3 \Rightarrow \alpha-\beta+\gamma=\frac{4}{3}
\end{aligned}$
solving equations (i), (ii), and (iii), we get
$\begin{aligned}
& \alpha=0, \beta=\frac{-4}{3}, \gamma=0 \\
& \therefore 5 \alpha-9 \beta+13 \gamma=12
\end{aligned}$