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 ${\mathrm{x}}^2-\left(3\sqrt{2}+2\sqrt{3}\right)\mathrm{x}+7+3\lambda \sqrt{3}=0$. If $\beta +\gamma =3\sqrt{2}$, then ${\left(\alpha +2\beta +\gamma \right)}^2$ is equal to

Given $\alpha \& \beta$ are roots of $x^2-4\lambda x+5=0$

So, $\alpha +\beta =4\lambda$ and $\alpha \beta =5$ .......(i)

And $\alpha \& \gamma$ are roots of $x^2-\left(3\sqrt{2}+2\sqrt{3}\right)x+\left(7+3\lambda \sqrt{3}\right)=0$

So, $\alpha +y=3\sqrt{2}+2\sqrt{3}$ and  $\alpha \gamma =7+3\lambda \sqrt{3}$ .........(ii)

Given, if $\beta +\gamma =3\sqrt{2}$

So, from equation (i) and (ii) we get, $\alpha =2\lambda +\sqrt{3}$ and 

$\beta =2\lambda -\sqrt{3}$, Now by product of roots we get, $4{\lambda }^2-3=5\Rightarrow \lambda =\sqrt{2}$

$\therefore {\left(\alpha +2\beta +\lambda \right)}^2= {\left(\alpha +\beta +\beta +\lambda \right)}^2={\left(4\lambda +3\sqrt{2}\right)}^2$

$={\left(4\sqrt{2}+3\sqrt{2}\right)}^2={\left(7\sqrt{2}\right)}^2=98$