II. (25 points) Given the quadratic function
$$
y=x^{2}+b x-c
$$

the graph passes through three points
$$
P(1, a), Q(2,3 a)(a \geqslant 3), R\left(x_{0}, y_{0}\right) .
$$

If the centroid of $\triangle P Q R$ is on the $y$-axis, find the minimum perimeter of $\triangle P Q R$.
Given points $P$ and $Q$ are on the graph of the quadratic function
$$
y=x^{2}+b x-c
$$

we have,
$$
\left\{\begin{array} { l } 
{ 1 + b - c = a , } \\
{ 4 + 2 b - c = 3 a }
\end{array} \Rightarrow \left\{\begin{array}{l}
b=2 a-3, \\
c=a-2 .
\end{array}\right.\right.
$$

Since the centroid of $\triangle P Q R$ lies on the $y$-axis, we get
$$
\begin{array}{l}
1+2+x_{0}=0 \Rightarrow x_{0}=-3 \\
\Rightarrow y_{0}=9-3 b-c=20-7 a .
\end{array}
$$

By translating $\triangle P Q R$ so that point $R$ moves to the origin, we obtain $\triangle P^{\prime} Q^{\prime} O$, where
$$
O(0,0), P^{\prime}(4,8 a-20), Q^{\prime}(5,10 a-20) \text {. }
$$

Let $10 a-20=5 t$.
Given $a \geqslant 3$, we know $t \geqslant 2$.
Also, $8 a-20=4 t-4$, thus
$$
P^{\prime}(4,4 t-4), Q^{\prime}(5,5 t) \text {. }
$$

Therefore, the perimeter of $\triangle P Q R$ is
$$
\begin{aligned}
f(t)= & \sqrt{4^{2}+(4 t-4)^{2}}+\sqrt{5^{2}+(5 t)^{2}}+ \\
& \sqrt{1+(t+4)^{2}}(t \geqslant 2) .
\end{aligned}
$$

It is easy to see that $f(t)$ is an increasing function for $t \in[2,+\infty)$.
Thus, $f(t)_{\min }=f(2)=4 \sqrt{2}+5 \sqrt{5}+\sqrt{37}$.
Therefore, the minimum perimeter of $\triangle P Q R$ is
$$
4 \sqrt{2}+5 \sqrt{5}+\sqrt{37} \text {. }
$$