(50 points) Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=$ $x^{3}+a x^{2}+b x+c$ have three 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 $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$.
Let $S=\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$, then
$$
\begin{aligned}
S & =\frac{27\left(\frac{2}{27} a^{3}-\frac{1}{3} a b+c\right)}{\lambda^{3}}=\frac{27 f\left(-\frac{1}{3} a\right)}{\lambda^{3}} \\
& =\frac{27\left(-\frac{1}{3} a-x_{1}\right)\left(-\frac{1}{3} a-x_{2}\right)\left(-\frac{1}{3} a-x_{3}\right)}{\left(x_{2}-x_{1}\right)^{3}} \\
& =\frac{-27\left(x_{1}+\frac{1}{3} a\right)\left(x_{2}+\frac{1}{3} a\right)\left(x_{3}+\frac{1}{3} a\right)}{\left(x_{2}-x_{1}\right)^{3}} .
\end{aligned}
$$
Let $\hat{\imath} u_{i}=x_{i}+\frac{a}{3}(i=1,2,3)$, then $u_{2}-u_{1}=x_{2}-x_{1}=\lambda$, $\stackrel{1}{.}\left(u_{1}+u_{2}\right)$, and $u_{1}+u_{2}+u_{3}=x_{1}+x_{2}+x_{3}+a=0$. Therefore, $u_{1}, u_{2}, u_{3}$ still satisfy the corresponding conditions, and
$$
S=\frac{-27 u_{1} u_{2} u_{3}}{\left(u_{2}-u_{1}\right)^{3}} .
$$

From $u_{3}=-\left(u_{1}+u_{2}\right)>\frac{u_{1}+u_{2}}{2}$, we get $u_{1}+u_{2}<0$, and let
$$
v_{1}=\frac{-u_{1}}{u_{2}-u_{1}}, v=v_{2}=\frac{u_{2}}{u_{2}-u_{1}},
$$

then $v_{1}+v_{2}=1, v_{1}, v_{2}$ are both greater than $0, v_{1}-v_{2}=\frac{u_{3}}{u_{2}-u_{1}}>0$,
$$
\begin{aligned}
S & =27 v_{1} v_{2}\left(v_{1}-v_{2}\right)=27 v(1-v)(1-2 v) \\
& =27 \sqrt{\left(v-v^{2}\right)^{2}(1-2 v)^{2}} \\
& =27 \sqrt{2\left(v-v^{2}\right)\left(v-v^{2}\right)\left(\frac{1}{2}-2 v+2 v^{2}\right)} \\
& \leqslant 27 \sqrt{2 \times\left(\frac{1}{6}\right)^{3}} \\
& =\frac{3}{2} \sqrt{3} .
\end{aligned}
$$

The equality holds when $v=\frac{1}{2}\left(1-\frac{\sqrt{3}}{3}\right)$, and the corresponding cubic equation is $x^{3}-\frac{1}{2} x+\frac{\sqrt{3}}{18}=0, \lambda=1\left(x_{1}=-\frac{1}{2}\left(1+\frac{\sqrt{3}}{3}\right), x_{2}=\frac{1}{2}\left(1-\frac{\sqrt{3}}{3}\right), x_{3}=\frac{\sqrt{3}}{3}\right)$.

In summary, the maximum value sought is $\frac{3}{2} \sqrt{3}$.