One, (40 points) Find all integer triples $(a, b, c)$ such that
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$

is a positive integer of the form $1729^{m}(m \in \mathbf{N})$.
Let $x=a-b, y=b-c, z=c-a$.
Then $x+y+z=0$, and
$$
x y z=2 \times 1729^{m}-4(m \in \mathbf{N}) \text {. }
$$

Notice,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
=0 .
\end{array}
$$

Therefore, when $m \geqslant 1$,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}=3 x y z \\
=6 \times 1729^{m}-12 \equiv 2(\bmod 7) .
\end{array}
$$

Also, $n^{3} \equiv 0,1,-1(\bmod 7)$, without loss of generality, assume
$$
x^{3} \equiv 0(\bmod 7), y^{3} \equiv z^{3} \equiv 1(\bmod 7) \text {. }
$$
$$
\begin{array}{l}
\text { Then } y+z \equiv 0(\bmod 7) \\
\Rightarrow y \equiv-z(\bmod 7) \\
\Rightarrow y^{3} \equiv-z^{3}(\bmod 7),
\end{array}
$$

Contradiction.
Thus, $m=0$, i.e.,
$$
x+y+z=0, x y z=-2 \text {, and } x 、 y 、 z \in \mathbf{Z} \text {. }
$$

Without loss of generality, assume $x \geqslant y \geqslant z$, then only $x=y=1, z=-2$. At this point, $(a, b, c)=(k+2, k+1, k)(k \in \mathbf{Z})$. Therefore, the three numbers are $k, k+1, k+2(k \in \mathbf{Z})$.