Three, (25 points) Find all positive integer triples $(a, b, c)$ that satisfy $a^{2}+2 b^{2}+3 c^{2}=2008$.

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Three, because $a^{2}+2 b^{2}+3 c^{2}=2008$, where $2 b^{2}$ and $2008$ are both even numbers, so $a^{2}+3 c^{2}$ is even, and more specifically, $a^{2}+c^{2}$ is even.
Therefore, $a$ and $c$ have the same parity.
If $a$ and $c$ are both odd, let $a=2 m-1, c=2 n-1$ (where $m, n$ are positive integers). Then
$$
\begin{array}{l}
(2 m-1)^{2}+2 b^{2}+3(2 n-1)^{2}=2008 \\
\Rightarrow 4 m(m-1)+2 b^{2}+12 n(n-1) \\
=2004 .
\end{array}
$$

Since 412004, we have
$4 \mid 2 b^{2} \Rightarrow 2 \mid b^{2}$, which means $b$ is even.
Let $b=2 k$ (where $k$ is a positive integer). Then
$$
\begin{array}{l}
4 m(m-1)+8 k^{2}+12 n(n-1) \\
=2004 . \\
\text { Since } 2 \mid m(m-1), 2 \mid n(n-1) \\
\Rightarrow 8 \mid 4 m(m-1), 8 \mid 12 n(n-1) .
\end{array}
$$

Thus, 8 divides the left side of equation (1), but it does not divide the right side of equation (1), leading to a contradiction.

Therefore, $a$ and $c$ cannot both be odd, meaning $a$ and $c$ must both be even.
Let $a=2 a_{1}, c=2 c_{1}$ (where $a_{1}, c_{1}$ are positive integers). Then $4 a_{1}^{2}+2 b^{2}+12 c_{1}^{2}=2008$,

which simplifies to $2 a_{1}^{2}+b^{2}+6 c_{1}^{2}=1004$.
Thus, $b^{2}$ is even, meaning $b$ is even.
Let $b=2 b_{1}$ (where $b_{1}$ is a positive integer). Then
$$
a_{1}^{2}+2 b_{1}^{2}+3 c_{1}^{2}=502 \text {. }
$$

Similarly, $a_{1}$ and $c_{1}$ have the same parity.
(1) When $a_{1}$ and $c_{1}$ are both odd, we have
$$
1^{2}+2 \times 1^{2}+3 c_{1}^{2} \leqslant 502 \text {. }
$$

Solving this, we get $c_{1}^{2} \leqslant 166 \frac{1}{2}<169$.
Thus, $c_{1}<13$, meaning $c_{1}=1,3,5,7,9,11$.
Substituting the values of $c_{1}$ into equation (2), we find
$$
\begin{array}{l}
\left(a_{1}, b_{1}, c_{1}\right) \\
=(7,15,1),(5,15,3),(11,3,11) . \\
\text { Therefore, }(a, b, c) \\
=(14,30,2),(10,30,6),(22,6,22) .
\end{array}
$$

Thus, $(a, b, c)$
(2) When $a_{1}$ and $c_{1}$ are both even,

Let $a_{1}=2 a_{2}, c_{1}=2 c_{2}$ (where $a_{2}, c_{2}$ are positive integers).
Then $4 a_{2}^{2}+2 b_{1}^{2}+12 c_{2}^{2}=502$, which simplifies to
$$
2 a_{2}^{2}+b_{1}^{2}+6 c_{2}^{2}=251,
$$

It is easy to see that $b_{1}$ is odd.
Thus, $2 \times 1^{2}+1^{2}+6 c_{2}^{2} \leqslant 251$.
Solving this, we get $c_{2}^{2} \leqslant \frac{124}{3}<49$.
Thus, $c_{2}<7$, meaning $c_{2}=1,2,3,4,5,6$.
Substituting the values of $c_{2}$ into equation (3), we get
$\left(a_{2}, b_{1}, c_{2}\right)=(1,15,2)$.
Thus, $(a, b, c)=(4,30,8)$.
In summary, the positive integer triples $(a, b, c)$ have four values:
$$
(a, b, c)=(14,30,2),(10,30,6),
$$