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 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 $\frac{2 a^{3} + 27 c - 9 a b}{\lambda^{3}}$.
Due to $f(x)=f(x)-f\left(x_{3}\right)$
$$
=\left(x-x_{3}\right)\left[x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b\right] \text {, }
$$

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 get
$$
\left(a+x_{3}\right)^{2}-4\left(x_{3}^{2}+a x_{3}+b\right)=\lambda^{2} \text {, }
$$

which is $3 x_{3}^{2}+2 a x_{3}+\lambda^{2}+4 b-a^{2}=0$. From (2) we get
$$
x_{3}=\frac{1}{3}\left(-a+\sqrt{4 a^{2}-12 b-3 \lambda^{2}}\right)
$$

and $4 a^{2}-12 b-3 \lambda^{2} \geqslant 0$. It is easy to see that $f(x)=x^{3}+a x^{2}+b x+c$
$$
=\left(x+\frac{a}{3}\right)^{3}-\left(\frac{a^{2}}{3}-b\right)\left(x+\frac{a}{3}\right)+\frac{2}{27} a^{3}+c-\frac{1}{3} a b \text {. }
$$

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

From (1) we get
$$
\begin{array}{l}
x_{3}+\frac{a}{3}=\frac{1}{3} \sqrt{4 a^{2}-12 b-3 \lambda^{2}} \\
=\frac{2 \sqrt{3}}{3} \sqrt{\frac{a^{2}}{3}-b-\frac{\lambda^{2}}{4}} .
\end{array}
$$

Let $p=\frac{a^{2}}{3}-b$. From (2) and (3) we know $p \geqslant \frac{\lambda^{2}}{4}$ and
$$
\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} \sqrt{p-\frac{\lambda^{2}}{4}}\left(p-\lambda^{2}\right) \text {. }
$$

Let $y=\sqrt{p-\frac{\lambda^{2}}{4}}$, then $y \geqslant 0$ and
$$
\begin{array}{l}
\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} y\left(y^{2}-\frac{3}{4} \lambda^{2}\right) . \\
\because y^{3}-\frac{3 \lambda^{2}}{4} y+\frac{\lambda^{3}}{4}=y^{3}-\frac{3 \lambda^{2}}{4} y-\left(\frac{\lambda}{2}\right)^{3}+\frac{3 \lambda^{2}}{4} \cdot \frac{\lambda}{2} \\
\quad=\left(y-\frac{\lambda}{2}\right)^{2}(y+\lambda) \geqslant 0, \\
\therefore \frac{1}{3} a b-\frac{2}{27} a^{3}-c \geqslant-\frac{\sqrt{3}}{18} \lambda^{3} .
\end{array}
$$

Thus, $2 a^{3}+27 c-9 a b \leqslant \frac{3 \sqrt{3}}{2} \lambda^{3}$. Therefore, $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}} \leqslant \frac{3 \sqrt{3}}{2}$.
By 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 assumption holds, 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 summary, the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$ is $\frac{3 \sqrt{3}}{2}$.