$$
\begin{array}{l}
\text { Three. (50 points) Given } a, b, c > 0 \text {, and } \\
a^{2}+b^{2}+c^{2}+a b c=4, p=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}, \\
s=a+b+c, t=2+a b c .
\end{array}
$$

Try to compare the sizes of $p$, $s$, and $t$, and find the maximum and minimum values of $p$, $s$, and $t$ respectively.
When $a=b=c=1$, $a^{2}+b^{2}+c^{2}+a b c=4, p=s=t=3$.
When $a=b=\frac{1}{2}$, according to the problem,
$\frac{1}{4}+\frac{1}{4}+c^{2}+\frac{1}{4} c=4$
$\Rightarrow 4 c^{2}+c-14=0$
$\Rightarrow c=\frac{7}{4}$ (discard the negative root).
At this time, $p>3>s>t$.
Guess: $p \geqslant 3 \geqslant s \geqslant t$.
Next, prove: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant 3 \geqslant a+b+c \geqslant 2+a b c$.
(1) Note that,
$$
\begin{array}{l}
4=a^{2}+b^{2}+c^{2}+a b c \\
\geqslant 4 \sqrt[4]{a^{2} b^{2} c^{2} a b c}=4 \sqrt[4]{(a b c)^{3}} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$

And $p=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant 3 \sqrt[3]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}}$
$$
=\frac{3}{\sqrt[3]{a b c}} \geqslant \frac{3}{\sqrt[3]{1}}=3 \text {. }
$$
(2) Note that,
$$
\begin{array}{l}
4=a^{2}+b^{2}+c^{2}+a b c \geqslant a^{2}+2 b c \text { then } 0 \geqslant\left(a^{2}-4\right)+b c(2+a) \\
=(2+a)(a-2+b c) .
\end{array}
$$

And $2+a>0$, so $0 \geqslant a-2+b c$.
Also, $a-1, b-1, c-1$ cannot all have the same sign, and at least two of them must have the same sign (let's assume $b-1, c-1$), then
$$
\begin{array}{l}
(b-1)(c-1) \geqslant 0 \\
\Rightarrow b c-b-c+1 \geqslant 0 \geqslant a-2+b c \\
\Rightarrow 3 \geqslant a+b+c=s .
\end{array}
$$
(3) From $0 \geqslant a-2+b c$
$$
\begin{array}{l}
\Rightarrow 2 \geqslant a+b c \\
\Rightarrow 2 a \geqslant a^{2}+a b c .
\end{array}
$$

Similarly, $2 b \geqslant b^{2}+a b c, 2 c \geqslant c^{2}+a b c$.
Adding the above three inequalities, we get
$$
\begin{array}{l}
2(a+b+c) \\
\geqslant\left(a^{2}+b^{2}+c^{2}+a b c\right)+2 a b c \\
=4+2 a b c .
\end{array}
$$

Thus, $s=a+b+c \geqslant 2+a b c=t$.
Therefore, $p \geqslant 3 \geqslant s \geqslant t$.
In summary, when and only when $a=b=c=1$,
$$
p_{\min }=3, s_{\max }=3, t_{\max }=3 .
$$