Let $S = \frac{2a^3 + 27c - 9ab}{\lambda^3}$, then
$$S = \frac{27\left(\frac{2}{27}a^3 - \frac{1}{3}ab + c\right)}{\lambda^3} = \frac{27f\left(-\frac{1}{3}a\right)}{\lambda^3} \\ = \frac{27\left(-\frac{1}{3}a - x_1\right)\left(-\frac{1}{3}a - x_2\right)\left(-\frac{1}{3}a - x_3\right)}{(x_2 - x_1)^3} \\ = \frac{-27\left(x_1 + \frac{1}{3}a\right)\left(x_2 + \frac{1}{3}a\right)\left(x_3 + \frac{1}{3}a\right)}{(x_2 - x_1)^3}.$$

Write $u_i = x_i + \frac{a}{3}$ ($i=1,2,3$), then $u_2 - u_1 = x_2 - x_1 = \lambda$, $u_3 > \frac{1}{2}(u_1 + u_2)$, and $u_1 + u_2 + u_3 = x_1 + x_2 + x_3 + a = 0$. So, $u_1$, $u_2$ and $u_3$ satisfy the corresponding conditions too, and
$$S = \frac{-27u_1 u_2 u_3}{(u_2 - u_1)^3}.$$
By $u_3 = -(u_1 + u_2) > \frac{u_1 + u_2}{2}$, we get $u_1 + u_2 < 0$. Consequently, at least one of $u_1$ and $u_2$ should be less than $0$. We may assume $u_1 < 0$.

If $u_2 < 0$, then $S < 0$.

If $u_2 > 0$, we suppose further
$$v_1 = \frac{-u_1}{u_2 - u_1}, \quad v = v_2 = \frac{u_2}{u_2 - u_1}.$$
Then $v_1 + v_2 = 1$, $v_1$ and $v_2$ are both greater than $0$ and $v_1 - v_2 = \frac{u_3}{u_2 - u_1} > 0$. So it follows that
$$\begin{aligned} S &= 27v_1v_2(v_1 - v_2) = 27v(1-v)(1-2v) \\ &= 27\sqrt{(v-v^2)^2(1-2v)^2} \\ &= 27\sqrt{2(v-v^2)(v-v^2)\left(\frac{1}{2}-2v+2v^2\right)} \\ &\le 27\sqrt{2 \times \left(\frac{1}{6}\right)^3} \\ &= \frac{3}{2}\sqrt{3}. \end{aligned}$$
The equality holds when $v = \frac{1}{2}\left(1 - \frac{\sqrt{3}}{3}\right)$. The corresponding cubic equation is $x^3 - \frac{1}{2}x + \frac{\sqrt{3}}{18} = 0$ and $\lambda = 1$ ($x_1 = -\frac{1}{2}(1 + \frac{\sqrt{3}}{3})$, $x_2 = \frac{1}{2}(1 - \frac{\sqrt{3}}{3})$ and $x_3 = \frac{\sqrt{3}}{3}$).

Consequently, the maximal value is $\frac{3}{2}\sqrt{3}$.