1. Use Vieta's formulas for both quadratic equations

For $x^2-4\lambda x+5=0,$ with roots $\alpha,\beta$: $\alpha+\beta=4\lambda, \qquad \alpha\beta=5.$

For $x^2-(3\sqrt2+2\sqrt3)x+7+3\lambda\sqrt3=0,$ with roots $\alpha,\gamma$: $\alpha+\gamma=3\sqrt2+2\sqrt3, \qquad \alpha\gamma=7+3\lambda\sqrt3.$

Also given: $\beta+\gamma=3\sqrt2.$

2. Find $\alpha$ using sum relations

Add the first two root sums: $\alpha+\beta=4\lambda$ $\alpha+\gamma=3\sqrt2+2\sqrt3$

Subtract the given relation $\beta+\gamma=3\sqrt2$ from their sum: $(\alpha+\beta)+(\alpha+\gamma)-(\beta+\gamma)=4\lambda+(3\sqrt2+2\sqrt3)-3\sqrt2$ $2\alpha=4\lambda+2\sqrt3$ $\alpha=2\lambda+\sqrt3.$

3. Express $\beta$ and $\gamma$ in terms of $\lambda$

From $\alpha+\beta=4\lambda$, $\beta=4\lambda-\alpha=4\lambda-(2\lambda+\sqrt3)=2\lambda-\sqrt3.$

From $\beta+\gamma=3\sqrt2$, $\gamma=3\sqrt2-\beta=3\sqrt2-(2\lambda-\sqrt3)=3\sqrt2-2\lambda+\sqrt3.$

4. Use the product relation $\alpha\beta=5$ to find $\lambda$

$\alpha\beta=(2\lambda+\sqrt3)(2\lambda-\sqrt3)=5$ $4\lambda^2-3=5$ $4\lambda^2=8$ $\lambda^2=2.$

Since $\lambda>0$, $\lambda=\sqrt2.$

Hence, $\alpha=2\sqrt2+\sqrt3, \qquad \beta=2\sqrt2-\sqrt3.$

And $\gamma=3\sqrt2-\beta=3\sqrt2-(2\sqrt2-\sqrt3)=\sqrt2+\sqrt3.$

5. Compute $\alpha+2\beta+\gamma$

$\alpha+2\beta+\gamma=(2\sqrt2+\sqrt3)+2(2\sqrt2-\sqrt3)+(\sqrt2+\sqrt3).$

Combine like terms:

• Coefficient of $\sqrt2$: $2+4+1=7$
• Coefficient of $\sqrt3$: $1-2+1=0$

So, $\alpha+2\beta+\gamma=7\sqrt2.$

Therefore, $(\alpha+2\beta+\gamma)^2=(7\sqrt2)^2=98.$

6. Comparison with stored answer

Derived answer = $98$.

This matches the stored correct answer.