Let $\alpha ,\beta$ be the roots of the equation $x^2-4\lambda x+5=0$ and $\alpha ,\gamma$ be the roots of the equation $x^2-(3\sqrt{2}+2\sqrt{3})x+7+3\lambda \sqrt{3}=0$. If $\beta +\gamma =3\sqrt{2}$, then $(\alpha +2\beta +\gamma {)}^2$ is equal to :

$\because \alpha ,\beta$ are roots of $x^2-4\lambda x+5=0$
$\therefore \alpha +\beta =4\lambda$ and $\alpha \beta =5$
Also, $\alpha ,\gamma$ are roots of

$$\begin{array}{rl} & x^2-(3\sqrt{2}+2\sqrt{3})x+7+3\sqrt{3}\lambda =0,\lambda >0\\ & \therefore \alpha +\gamma =3\sqrt{2}+2\sqrt{3},\alpha \gamma =7+3\sqrt{3}\lambda \end{array}$$

$\because \alpha$ is common root

$$\begin{array}{ll}\therefore & {\alpha }^2-4\lambda \alpha +5=0\\ & \text{ and }{\alpha }^2-(3\sqrt{2}+2\sqrt{3})\alpha +7+3\sqrt{3}\lambda =0\end{array}$$

From (i) - (ii): we get $\alpha =\frac{2+3\sqrt{3}\lambda }{3\sqrt{2}+2\sqrt{3}-4\lambda }$

$$\because \beta +\gamma =3\sqrt{2}$$

$$\begin{gathered} \Rightarrow 4\lambda -\alpha +3\sqrt{2}+2\sqrt{3}-\alpha =3\sqrt{2} \\ \therefore 4\lambda +3\sqrt{2}+2\sqrt{3}-2\alpha =3\sqrt{2} \end{gathered}$$

$$\begin{gathered} \Rightarrow 3\sqrt{2}=4\lambda +3\sqrt{2}+2\sqrt{3}-\frac{4+6\sqrt{3}\lambda }{3\sqrt{2}+2\sqrt{3}-4\lambda } \\ \Rightarrow 3\sqrt{2}(3\sqrt{2}+2\sqrt{3}-4\lambda )={\left(3\sqrt{2}+2\sqrt{3}\right)}^2-{\left(4\lambda \right)}^2-\left(4+6\sqrt{3}\lambda \right) \\ \Rightarrow 18+6\sqrt{6}-12\sqrt{2}\lambda =18+12+6\sqrt{6}-16{\lambda }^2-4-6\sqrt{3}\lambda \\ \Rightarrow 16{\lambda }^2+6\sqrt{3}\lambda -12\sqrt{2}\lambda -8-6\sqrt{6}=0 \end{gathered}$$

$$\begin{array}{rl} & \Rightarrow 8{\lambda }^2+3(\sqrt{3}-2\sqrt{2})\lambda -4-3\sqrt{6}=0\\ & \therefore \lambda =\frac{6\sqrt{2}-3\sqrt{3}\pm \sqrt{9(11-4\sqrt{6})+32(4+3\sqrt{6})}}{16}\end{array}$$

$$\begin{array}{r}\therefore \lambda =\sqrt{2}\\ \therefore (\alpha +2\beta +\gamma {)}^2& =(\alpha +\beta +\beta +\gamma {)}^2\\ & =(4\sqrt{2}+3\sqrt{2}{)}^2\\ & =(7\sqrt{2}{)}^2\\ & =98\end{array}$$