Let $x=a-1$, $y=b-1$, and $z=c-1$. Then we need to satisfy $$xyz \text{ } | \text{ } (xz+yz+zx+x+y+z)$$ where $1\le x < y < z$.

Let $xy+yx+zx+x+y+z = kxyz$, where $k$ is an integer. This can also be written as: $$k=\frac{xy+yz+zx+x+y+z}{xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$$

We are going to show $k$ can only take $1$ or $2$.

For convenient, let $k=f(x, y, z)$. It is clearly from the expression above that $k=f(x, y, z)$ is a decreasing function with respect to each of the variables $x$, $y$, and $z$. Because $x$, $y$, and $z$ are all distinct, we have $$k = f(x,y,z) \le f(1, 2, 3)= \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 1} < 3$$

Therefore $k$ can only take $1$ or $2$.

Meanwhile, we also note $f(3, 4, 5) = \frac{59}{60} < 1$. This means $x < 3$, i.e. $x$ can only be $1$ or $2$.

When $x = 2$, $k=f(2, y, z) \le f(2, 3, 4) = \frac{35}{24} < 2$. Thus, we necessarily have $k=1$ in this situation. By now, we find we only need to consider three cases:

• $x=1$, $k=1$
• $x=1$, $k=2$
• $x=2$, $k=1$

For equation: $xy+yz+zx+x+y+z=kxyz$

Case (1) is equivalent to $y+yz+z+1+y+z=yz$ or $$1+2(y+z)=0$$ It has no solution.

Case (2) is equivalent to $y+yz+z+1+y+z=2yz$ or $$1+2(y+z)=yz$$ It can be rewritten as $(y-2)(z-2) = 5$. Because $ y< z $, we obtain a unique solution in this case $y=3$ and $z=7$.

Case (3) is equivalent to $2y + yz + 2z + 2 + y + z = yz$, or $$2+3(y+z)=yz$$ It can be rewritten as $(y-3)(z-3)=11$, which has a unique solution $y=4$ and $z=4$.

Hence, we conclude there are two solutions $(x, y, z) = (1, 3, 7), (2, 4, 14)$. Accordingly there are two solutions to the original equation: $$(a, b, c) = \boxed{(2, 4, 8), (3, 5, 15)}$$