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

<p>$\because$ $\alpha$, $\beta$ are roots of x² $-$ 4$\lambda$x + 5 = 0</p> <p>$\therefore$ $\alpha$ + $\beta$ = 4$\lambda$ and $\alpha$$\beta$ = 5</p> <p>Also, $\alpha$, $\gamma$ are roots of</p> <p>${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0$</p> <p>$\therefore$ $\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3$, $\alpha \gamma = 7 + 3\sqrt 3 \lambda$</p> <p>$\because$ $\alpha$ is common root</p> <p>$\therefore$ ${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$ ....... (i)</p> <p>and ${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$ ...... (ii)</p> <p>From (i) - (ii) : we get $\alpha = {{2 + 3\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$</p> <p>$\because$ $\beta + \gamma = 3\sqrt 2$</p> <p>$\therefore$ $4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2$</p> <p>$$ \Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$$</p> <p>$\Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$</p> <p>$\therefore$ $$\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$$</p> <p>$\therefore$ $\lambda = \sqrt 2$</p> <p>$\therefore$ ${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$</p> <p>$= {(4\sqrt 2 + 3\sqrt 2 )^2}$</p> <p>$= {(7\sqrt 2 )^2} = 98$</p>