$x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+(7+3 \lambda \sqrt{3})=0\left\langle_{\gamma}^{\alpha}\right.$
$\alpha+\beta=4 \lambda$
$\alpha+\gamma=3 \sqrt{2}+2 \sqrt{3}$
$\begin{array}{lll} \beta+\lambda=3 \sqrt{2} \alpha \gamma=7+3 \lambda \sqrt{3} \\ \therefore \quad \alpha=2 \lambda+\sqrt{3} \alpha \beta=5 \\ \beta=2 \lambda-\sqrt{3} 4 \lambda^{2}=8 \Rightarrow \lambda=\sqrt{2} \\ \therefore \quad (\alpha+2 \beta+\lambda)^{2}=(4 \alpha+3 \sqrt{2})^{2}=(7 \sqrt{2})^{2}=98\end{array}$
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