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 $1729^{m}(m \in \mathbb{N})$.
II. Solution: Let $x=a-b, y=b-c, z=c-a$, then $x+y+z=0, xyz=2 \cdot 1729^{m}-4 (m \in \mathbb{N})$.
From $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)=0$, we know that when $m \geq 1$, $x^{3}+y^{3}+z^{3}=3xyz=6 \cdot 1729^{m}-12 \equiv 2 \pmod{7}$. For any integer $k$, we have $k^{3} \equiv 0, \pm 1 \pmod{7}$, so among $x^{3}, y^{3}, z^{3}$, there must be two numbers that are $1 \pmod{7}$ and one number that is $0 \pmod{7}$. We assume without loss of generality that $x^{3} \equiv 0 \pmod{7}$, $y^{3} \equiv z^{3} \equiv 1 \pmod{7}$, hence $x \equiv 0 \pmod{7}$. Since $x+y+z=0$, it follows that $y+z \equiv 0 \pmod{7}$, i.e., $1 \equiv y^{3} \equiv (-z)^{3} \equiv -z^{3} \equiv -1 \pmod{7}$, which is a contradiction! Therefore, it must be that $m=0$.

Thus, $x+y+z=0, xyz=2$. Assuming $x \geq y \geq z$, and since $x, y, z \in \mathbb{Z}$, the only possibility is $x=1, y=1, z=-2$. This corresponds to $(a, b, c)=(k+2, k+1, k)$, so the integer triples $(a, b, c)$ are $(p, q, r)$, where $p, q, r$ are a permutation of $k, k+1, k+2 (k \in \mathbb{Z})$.