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 ______

Given,

$$x^2-4\lambda x+5=0$$

has roots $$\alpha,\beta$$.

Hence,

$$\alpha+\beta=4\lambda,\qquad \alpha\beta=5$$

Also,

$$x^2-(3\sqrt2+2\sqrt3)x+7+3\lambda\sqrt3=0$$

has roots $$\alpha,\gamma$$.

Therefore,

$$\alpha+\gamma=3\sqrt2+2\sqrt3$$

Given,

$$\beta+\gamma=3\sqrt2$$

Now,

$$2\alpha+(\beta+\gamma)=(\alpha+\beta)+(\alpha+\gamma)$$

Substituting the values,

$$2\alpha+3\sqrt2=4\lambda+3\sqrt2+2\sqrt3$$

$$2\alpha=4\lambda+2\sqrt3$$

$$\alpha=2\lambda+\sqrt3$$

Since $$\alpha$$ is a root of

$$x^2-4\lambda x+5=0$$

we get

$$\alpha^2-4\lambda\alpha+5=0$$

Substituting $$\alpha=2\lambda+\sqrt3$$,

$$\left(2\lambda+\sqrt3\right)^2-4\lambda\left(2\lambda+\sqrt3\right)+5=0$$

$$4\lambda^2+4\lambda\sqrt3+3-8\lambda^2-4\lambda\sqrt3+5=0$$

$$-4\lambda^2+8=0$$

$$\lambda^2=2$$

Taking $$\lambda=\sqrt2$$,

$$\alpha+\beta=4\lambda=4\sqrt2$$

Now,

$$\alpha+2\beta+\gamma=(\alpha+\beta)+(\beta+\gamma)$$

$$=4\sqrt2+3\sqrt2$$

$$=7\sqrt2$$

Therefore,

$$\left(\alpha+2\beta+\gamma\right)^2=(7\sqrt2)^2=98$$

Hence,

$$\boxed{98}$$