Three. (20 points) Let real numbers $p, q, r$ satisfy: there exists $a$ which is one of $p, q, r$, and the other two are exactly the two real roots of the equation
$$
x^{2}+(a-3) x+a^{2}-3 a=0
$$

Find the minimum possible value of $p^{3}+q^{3}+r^{3}$.
Three, let $a=p$, then $q, r$ are exactly the two roots of the equation. By Vieta's formulas, we know $p+q+r=3$ and $qr=p^2-3p$.
From $\Delta=(p-3)^2-4(p^2-3p) \geqslant 0$, we know $-1 \leqslant p \leqslant 3$.
$$
\begin{array}{l}
\text { Then } p^2+q^2+r^2=p^2+(q+r)^2-2qr \\
=p^2+(3-p)^2-2(p^2-3p)=9 \text {. }
\end{array}
$$

Thus, $p^3+q^3+r^3$
$$
\begin{array}{l}
=3pqr+(p+q+r)[p^2+q^2+r^2-p(q+r)-qr] \\
=3p(p^2-3p)+3[9-p(3-p)-(p^2-3p)] \\
=3p^3-9p^2+27,
\end{array}
$$

where $-1 \leqslant p \leqslant 3$.
For $f(p)=3p^3-9p^2+27$, taking the derivative gives
$$
f'(p)=9p^2-18p=9p(p-2) \text {. }
$$

Thus, $f(p)$ is increasing on $[2,3]$, decreasing on $[0,2]$, and increasing on $[-1,0]$.
Therefore, the minimum value of $f(p)$ on $[-1,3]$ is
$$
\min \{f(-1), f(2)\}=15 \text {. }
$$

Hence, the minimum value of $p^3+q^3+r^3$ is 15.
When $p=2, q=2, r=-1$, the minimum value can be achieved and the conditions are satisfied.