$$\because$$ $$\alpha$$, $$\beta$$ are roots of x² $$-$$ 4$$\lambda$$x + 5 = 0

$$\therefore$$ $$\alpha$$ + $$\beta$$ = 4$$\lambda$$ and $$\alpha$$$$\beta$$ = 5

Also, $$\alpha$$, $$\gamma$$ are roots of

$${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0$$

$$\therefore$$ $$\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3 $$, $$\alpha \gamma = 7 + 3\sqrt 3 \lambda $$

$$\because$$ $$\alpha$$ is common root

$$\therefore$$ $${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$$ ....... (i)

and $${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$$ ...... (ii)

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

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

$$\therefore$$ $$4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2 $$

$$ \Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$$

$$ \Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$$

$$\therefore$$ $$\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$$

$$\therefore$$ $$\lambda = \sqrt 2 $$

$$\therefore$$ $${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$$

$$ = {(4\sqrt 2 + 3\sqrt 2 )^2}$$

$$ = {(7\sqrt 2 )^2} = 98$$