I4.1 If $\frac{6 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}=3 \sqrt{\alpha}-6$, determine the value of $\alpha$.
$\begin{array}{l}\frac{6 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}} \cdot \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=3 \sqrt{\alpha}-6 \\ 6 \sqrt{3} \cdot \frac{3 \sqrt{2}-2 \sqrt{3}}{18-12}=3 \sqrt{\alpha}-6 \\ 3 \sqrt{6}-6=3 \sqrt{\alpha}-6 \\ \alpha=6 \\\end{array}$