One, (20 points) Given $m>0$, and $m$ is a rational number, $\sqrt{m}$ is an irrational number. Find all positive integers $a, b, c$, such that $\frac{\sqrt{m} a+b}{\sqrt{m} b+c}$ is a rational number, and $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=3$.
Given that $\sqrt{m} b - c$ is an irrational number, therefore, $\sqrt{m} b - c \neq 0$.
$$
\begin{array}{l}
\text { Hence } \frac{\sqrt{m} a + b}{\sqrt{m} b + c} = \frac{(\sqrt{m} a + b)(\sqrt{m} b - c)}{m b^2 - c^2} \\
= \frac{(m a b - b c) + \sqrt{m}(b^2 - a c)}{m b^2 - c^2}.
\end{array}
$$

Since the above expression is a rational number, we have
$$
b^2 - a c = 0.
$$

Then $a^2 + b^2 + c^2 = (a + b + c)^2 - 2 a b - 2 b c - 2 c a$
$$
\begin{array}{l}
= (a + b + c)^2 - 2(a b + b c + b^2) \\
= (a + b + c)(a - b + c).
\end{array}
$$

Thus, $\frac{a^2 + b^2 + c^2}{a + b + c} = a - b + c$. Therefore,
$$
a - b + c = 3.
$$

From equations (1) and (2), we know that $a$ and $c$ are the two positive integer roots of the quadratic equation in $x$
$$
x^2 - (b + 3)x + b^2 = 0.
$$

Thus,
$$
\Delta = [-(b + 3)]^2 - 4 b^2 \geqslant 0.
$$

Solving this, we get $b \leqslant 3$. Therefore, $b$ can only take the values $1, 2, 3$.
When $b = 1$, equation (3) has no integer roots;
When $b = 2$, the roots of equation (3) are $x_1 = 1, x_2 = 4$, so $a = 1, c = 4$ or $a = 4, c = 1$;
When $b = 3$, the roots of equation (3) are $x_3 = x_4 = 3$, so $a = c$.
Therefore, all the positive integer solutions $(a, b, c)$ are
$$
(1, 2, 4), (4, 2, 1), (3, 3, 3).
$$