Example 2 Real numbers $a, b, c$ and a positive number $\boldsymbol{\lambda}$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real numbers $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$;

Find: $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$'s maximum value. (2002 National High School Mathematics League Additional Question)
Let $x_{1}=m-\frac{\lambda}{2}, x_{2}=m+\frac{\lambda}{2}, x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)=m$, let $x_{3}=m+t,(\lambda >0, t>0)$.

By Vieta's formulas, we have
$$\left\{\begin{array}{l}
a=-\left(x_{1}+x_{2}+x_{3}\right)=-(3 m+t) \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=3 m^{2}+2 m t-\frac{\lambda^{2}}{4} \\
c=-x_{1} x_{2} x_{3}=-m^{3}-m^{2} t+\frac{\lambda^{2}}{4} m+\frac{\lambda^{2}}{4} t
\end{array}\right.$$

Then $2 a^{3}-9 a b=a\left(2 a^{2}-9 b\right)=27 m^{3}+27 m^{2} t-\frac{27}{4} m \lambda^{2}-\frac{9}{4} \lambda^{2} t-2 t^{3}, 27 c=$ $-27 m^{3}-27 m^{2} t+\frac{27}{4} m \lambda^{2}+\frac{27}{4} \lambda^{2} t$

Thus, $s=\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{9 \lambda^{2} t-4 t^{3}}{2 \lambda^{3}}$ (to achieve the maximum value, $9 \lambda^{2}-4 t^{2}$ should be positive) $=\frac{1}{4 \sqrt{2} \lambda^{3}} \sqrt{\left(9 \lambda^{2}-4 t^{2}\right)\left(9 \lambda^{2}-4 t^{2}\right) \cdot 8 t^{2}} \leqslant$
$$\frac{1}{4 \sqrt{2} \lambda^{3}} \sqrt{\left[\frac{\left(9 \lambda^{2}-4 t^{2}\right)+\left(9 \lambda^{2}-4 t^{2}\right)+8 t^{2}}{3}\right]^{3}}=\frac{3 \sqrt{3}}{2} .$$

When and only when $9 \lambda^{2}-4 t^{2}=8 t^{2} \Rightarrow t=\frac{\sqrt{3}}{2} \lambda$ (at this point $\left.9 \lambda^{2}-4 t^{2}>0\right)$, if we set $m=0, \lambda =2$, then $x_{1}=-1, x_{2}=1, x_{3}=\sqrt{3}$. It can be seen that when $a=-c=-\sqrt{3}, b=-1$, the maximum value $\frac{3 \sqrt{3}}{2}$ can be achieved.