Let's determine all integers $a, b, c$ such that $1<a<b<c$, and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$.
The parity of $a, b, c$ is the same. If there were both even and odd numbers among the three, then $(a-1)(b-1)(c-1)$ would be even, while $abc-1$ would be odd, and an even number cannot be a divisor of an odd number.

Let

$$
h=\frac{abc-1}{(a-1)(b-1)(c-1)}
$$

that is,

$$
abc-1=h(a-1)(b-1)(c-1)
$$

Since $(abc-1)-(a-1)(b-1)(c-1)=a(b-1)+b(c-1)+c(a-1)>0$, clearly $h>1$; on the other hand,

$$
h<\frac{abc}{(a-1)(b-1)(c-1)}=\left(1+\frac{1}{a-1}\right)\left(1+\frac{1}{b-1}\right)\left(1+\frac{1}{c-1}\right)
$$

We will show that $a<4$. If $a \geq 4$, then (since $a, b, c$ have the same parity) $b \geq a+2 \geq 6$ and $c \geq b+2 \geq 8$, so by (2),

$$
h<\left(1+\frac{1}{4-1}\right)\left(1+\frac{1}{6-1}\right)\left(1+\frac{1}{8-1}\right)=\frac{64}{35}<2
$$

This is a contradiction because $h$ is an integer and greater than 1.

We then break the problem into two cases based on whether $a$ is 2 or 3.

I. If $a=2$, then $abc-1$ is odd, so $h$ cannot be even. Also, $b \geq 4$ and $c \geq 6$, so by (2),

$$
h<\left(1+\frac{1}{2-1}\right)\left(1+\frac{1}{4-1}\right)\left(1+\frac{1}{6-1}\right)=\frac{16}{5}<4
$$

The only possible value for $h$ is 3. Substituting the assumed $a=2$ and the obtained $h=3$ into (1):

$$
2bc-1=3(b-1)(c-1)
$$

Rearranging and factoring:

$$
bc-3b-3c+4=0; \quad (b-3)(c-3)=5
$$

Since both factors on the left are positive, and the second factor is the larger, this is satisfied precisely when $b-3=1$ and $c-3=5$, that is,

$$
a=2, b=4, c=8
$$

II. If $a=3$, then $b \geq 5, c \geq 7$ and

$$
h<\left(1+\frac{1}{3-1}\right)\left(1+\frac{1}{5-1}\right)\left(1+\frac{1}{7-1}\right)=\frac{35}{16}<3
$$

so $h=2$. In this case, (1) becomes:

$$
3bc-1=4(b-1)(c-1)
$$

Rearranging and factoring:

$$
\begin{aligned}
bc-4b-4c+5 & =0 \\
(b-4)(c-4) & =11
\end{aligned}
$$

Since both factors are positive, and the second is the larger, this is satisfied precisely when $b-4=1$ and $c-4=11$, that is,

$$
a=3, \quad b=5, \quad c=15
$$