$$
\begin{array}{c}
\text { Six, let } f\left(x_{1}, x_{2}, x_{3}\right)=-2\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}\right)+3\left[x_{1}^{2}\right. \\
\left.\cdot\left(x_{2}+x_{3}\right)+x_{2}^{2}\left(x_{3}+x_{1}\right)+x_{3}^{2}\left(x_{1}+x_{2}\right)\right]-12 x_{1} x_{2} x_{3} \text {. }
\end{array}
$$

For any real numbers $r, s, t$, let
$$
g(r, s, t)=\max _{t \leqslant x_{3} \leqslant t+2}\left|f\left(r, r+2, x_{3}\right)+s\right| .
$$

Find the minimum value of the function $g(r, s, t)$.
Let $x=x_{3}-(r+1)$, then
$f\left(r, r+2, x_{3}\right)$
$=-2\left[r^{3}+(r+2)^{3}+(x+r+1)^{3}\right]$
$+3\left[r^{2}(x+2 r+3)+(r+2)^{2}(x+2 r+1)\right.$
$\left.+(x+r+1)^{2}(2 r+2)\right]-12 r(r+2)(x+r+1)$
$=-2 x^{3}+18 x$.
Let $a=t-(r+1)$. Then
$g(r, s, t)=\max _{u \leqslant x \leqslant a+2} 1-2 x^{3}+18 x+s \mid$.
Since $\max _{a \leqslant x \leqslant a+2} 1-2 \leqslant x^{3}+18 x+s \mid$
$\geqslant \frac{1}{2}\left[\max _{a \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x+s\right)-\min _{a \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x+s\right)\right]$
$=\frac{1}{2}\left[\max _{a \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x\right)-\min _{a \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x\right)\right]$.
Therefore, $g(r, s, t)$
$\geqslant \frac{1}{2}\left[\max _{u \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x\right)-\min _{u \leqslant x \leqslant a+2}\left(-2 x^{3}+18 x\right)\right]$. (1)
Let $P(x)=-2 x^{3}+18 x$. For any $x_0$;
When $-\sqrt{3} \leqslant x_0 \leqslant \sqrt{3}$, $P(x)=P(\dot{x})$ has three distinct real roots $x_{1} < x_{2} < x_{3}$, and $P(x_1) = P(x_2) = P(x_3)$. Since $P(x)$ is strictly increasing on $\left[-\sqrt{3}, x_{2}\right]$, we have $P(-\sqrt{3}) < P(x_2) = P(x_1) > P(a)$.
Thus, we can obtain
$W(a)=P(a+2)-P(-\sqrt{3})$
$\geqslant P\left(x_{1}\right)-P(-\sqrt{3})=W\left(x_{1}\right)$.
Therefore, when $-2-\sqrt{3} \leqslant a \leqslant -\sqrt{3}$, we have $W(a) \geqslant 12 \sqrt{3}-\frac{40 \sqrt{6}}{9}$, and when $a \geqslant -\sqrt{3}$, we have $W(a) \geqslant 24 \sqrt{3}-16 \geqslant 12 \sqrt{3}-\frac{40 \sqrt{6}}{9}$. Thus,
$\min _{a \in \mathbb{R}} W(a)=W\left(x_{1}\right)=12 \sqrt{3}-\frac{40 \sqrt{6}}{9}$.
From (1), we know that $g(r, s, t) \geqslant \frac{1}{2} W\left(x_{1}\right)=6 \sqrt{3}-\frac{20 \sqrt{6}}{9}$.
From the above proof, for any real number $r$,
$t=x_{1}+(r+1)=-\frac{2 \sqrt{6}}{3}-1+(r+1)$,
$s=-\frac{1}{2}\left[\max _{x_{1} \leqslant x \leqslant x_{1}+2} P(x)+\min _{x_{1} \leqslant x \leq x_{1}+2} P(x)\right]$,
it is easy to see that $g(r, s, t)=-\frac{1}{2} W\left(x_{1}\right)=6 \sqrt{3}-\frac{20 \sqrt{6}}{9}$.
Therefore, $\min _{r, \cdots \in \mathbb{R}} g(r, s, t)=6 \sqrt{3}-\frac{20 \sqrt{6}}{9}$.