II. (50 points)
Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b x+c$ have 3 real roots $x_{1}, x_{2}$, $x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.
Find the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$.
II. (50 points)
Since $f(x)=f(x)-f\left(x_{3}\right)$
$$
=\left(x-x_{3}\right)\left[x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b\right] \text {. }
$$

Therefore, $x_{1}, x_{2}$ are the roots of the equation $x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b=0$. From (1), we have
$$
\left(a+x_{3}\right)^{2}-4\left(x_{3}^{2}+a x_{3}+b\right)=\lambda^{2},
$$

which simplifies to
$$
3 x_{3}^{2}+2 a x_{3}+\lambda^{2}+4 b-a^{2}=0 \text {. }
$$

From (2), we get
$$
\begin{array}{c}
x_{3}=\frac{1}{3}\left[-a+\sqrt{4 a^{2}-12 b-3 \lambda^{2}}\right] \\
4 a^{2}-12 b-3 \lambda^{2} \geq 0
\end{array}
$$

and
$$
4 a^{2}-12 b-3 \lambda^{2} \geq 0
$$

It is easy to see that $f(x)=x^{3}+a x^{2}+b x+c$
$$
=\left(x+\frac{a}{3}\right)^{3}-\left(\frac{a^{2}}{3}-b\right)\left(x+\frac{a}{3}\right)+\frac{2}{27} a^{3}+c-\frac{1}{3} a b .
$$

From $f\left(x_{3}\right)=0$, we get $\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\left(x_{3}+\frac{a}{3}\right)^{3}-\left(\frac{a^{2}}{3}-b\right)\left(x_{3}+\frac{a}{3}\right)$

From (1), we have $x_{1}+\frac{a}{3}=\frac{1}{3} \sqrt{4 a^{2}-12 b-3 \lambda^{2}}=\frac{2 \sqrt{3}}{3} \sqrt{\frac{a^{2}}{3}-b-\frac{\lambda^{2}}{4}}$.
Let $p=\frac{a^{2}}{3}-b$, from (2) and (3) we know $p \geqslant \frac{\lambda^{2}}{4}$, so
$$
\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} \sqrt{p-\frac{\lambda^{2}}{4}}\left(p-\lambda^{2}\right) .
$$

Let $y=\sqrt{p-\frac{\lambda^{2}}{4}}$, then $y \geqslant 0$ and $\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} y\left(y^{2}-\frac{3}{4} \lambda^{2}\right)$.
Since $y^{3}-\frac{3 \lambda^{2}}{4} y+\frac{\lambda^{3}}{4}=y^{3}-\frac{3 \lambda^{2}}{4} y=\left(\frac{\lambda}{2}\right)^{3}+\frac{3 \lambda^{2}}{4} \cdot \frac{\lambda}{2}$
$$
\begin{array}{l}
=\left(y-\frac{\lambda}{2}\right)\left(y^{2}+\frac{\lambda}{2} y+\frac{\lambda^{2}}{4}-\frac{3 \lambda^{2}}{4}\right) \\
=\left(y-\frac{\lambda}{2}\right)^{2}(y+\lambda) \geqslant 0,
\end{array}
$$

Therefore,
$$
\frac{1}{3} a b-\frac{2}{27} a^{3}-c \geqslant-\frac{\sqrt{3}}{18} \lambda^{3} \text {. }
$$

Thus,
$$
2 a^{3}+27 c-9 a b \leqslant \frac{3 \sqrt{3}}{2} \lambda^{3} .
$$

Hence,
$$
\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}} \leqslant \frac{3 \sqrt{3}}{2} \text {. }
$$

Taking $a=2 \sqrt{3}, b=2, c=0, \lambda=2$, then $f(x)=x^{3}+2 \sqrt{3} x^{2}+2 x$ has roots $-\sqrt{3}-1, -\sqrt{3}+1, 0$. Clearly, the assumptions hold, and
$$
\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{1}{8}(48 \sqrt{3}-36 \sqrt{3})=\frac{3 \sqrt{3}}{2} .
$$

In conclusion, the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$ is $\frac{3 \sqrt{3}}{2}$.