Given,
$$
\begin{aligned}
& A=(\alpha, 1,2 \alpha) \\
& B=(3,1,2) \\
& \mathbf{C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{aligned}
$$
and
Now, $\quad \mathbf{A B}=(3-\alpha) \hat{\mathbf{i}}+(2-2 \alpha) \hat{\mathbf{k}}$
$$
\begin{aligned}
\therefore & \mathbf{A B} \times \mathbf{C}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3-\alpha & 0 & 2-2 \alpha \\
4 & -1 & 3
\end{array}\right| \\
& =\hat{\mathbf{i}}(2-2 \alpha)-\hat{\mathbf{j}}(9-3 \alpha-8+8 \alpha)+\hat{\mathbf{k}}(-3+\alpha) \\
& =(2-2 \alpha) \hat{\mathbf{i}}+(5 \alpha-1) \hat{\mathbf{j}}+(\alpha-3) \hat{\mathbf{k}}
\end{aligned}
$$

Comparing it with $6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$, then
$$
\begin{aligned}
& 2-2 \alpha=6,5 \alpha-1=9 \text { and } \alpha-3=-5 \\
& 2 \alpha=-4 \\
& \Rightarrow \quad \alpha=-2 \\
& \therefore \quad \alpha^2+\alpha+5=(-2)^2-2+5 \\
& =4-2+5=7 \\
&
\end{aligned}
$$