Three. (25 points) Find the value of the real number $a$ such that the graph of the function
$$
f(x)=(x+a)(|x-a+1|+|x-3|)-2 x+4 a
$$
is centrally symmetric.
For the convenience of narration, use
$$
\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} 、 \min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}
$$

to represent the maximum and minimum numbers among $a_{1}, a_{2}, \cdots, a_{n}$, respectively.
When $x \leqslant \min \{a-1,3\}$, we have
$$
\begin{array}{l}
f(x)=(x+a)(-2 x+a+2)-2 x+4 a \\
=-2 x^{2}-a x+a^{2}+6 a .
\end{array}
$$

When $x \geqslant \max \{a-1,3\}$, we have
$$
\begin{array}{l}
f(x)=(x+a)(2 x-a-2)-2 x+4 a \\
=2 x^{2}+(a-4) x-a^{2}+2 a .
\end{array}
$$

When $\min \{a-1,3\}<x<\max \{a-1,3\}$, $f(x)$ is a linear function.

It can be seen that the graph of the function $f(x)$ consists of two parabolic arcs on the left and right sides, and a line segment in the middle. The graph of the function $f(x)$ is a centrally symmetric figure if and only if its center of symmetry is the midpoint $M$ of the middle line segment, and the vertices $A$ and $B$ of the parabolic arcs on the left and right sides are also symmetric about point $M$.
Since $x_{M}=\frac{(a-1)+3}{2}=\frac{a+2}{2}$,
$$
x_{A}=-\frac{a}{4}, x_{B}=-\frac{a-4}{4},
$$

we have $\frac{x_{A}+x_{B}}{2}=x_{M}$, i.e.,
$$
-\frac{a+(a-4)}{8}=\frac{a+2}{2} \text {. }
$$

Solving for $a$, we get $a=-\frac{2}{3}$.
Upon verification, $a=-\frac{2}{3}$ satisfies the requirements.
(Liu Kangning, Xi'an Tieyi Middle School, Xi'an, Shaanxi Province, 710054; Lü Jianhui, Xingping City Educational Research Office, Shaanxi Province, 713100)