Solve the following equation in the set of positive integers:

$$
\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=2 \text {. }
$$
The roles of the numbers $a, b, c$ are symmetric, so we can assume that the relation $a \leq b \leq c$ holds. The solutions obtained this way can then be permuted to get all possible solutions.

Let's examine the smallest number. If $a=1$, then

$$
\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=2\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)>2
$$

thus, necessarily $a \geq 2$. However, if $a \geq 4$, then

$$
\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \leq\left(1+\frac{1}{4}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{4}\right)=\left(\frac{5}{4}\right)^{3}=\frac{125}{64}<2,
$$

which also does not provide a solution. Therefore, only $a=2$ or $a=3$ are possible.

In the case of $a=2$, the equation can be rewritten as:

$$
\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{4}{3}, 3(b+1)(c+1)=4 b c, 3 b+3 c+3=b c, b=\frac{3+3 c}{c-3}=\frac{3(c-3)+12}{c-3}=3+\frac{12}{c-3} .
$$

Since $b$ is an integer and $c \geq 2$, the possible values of $c-3$ are the divisors of 12 and not less than -1: $c-3=-1,1,2,3,4,6$, 12. The corresponding $(b, c)$ pairs are:

$$
(-9,2),(15,4),(9,5),(7,6),(6,7),(5,9),(4,15)
$$

Among these, only $(6,7),(5,9),(4,15)$ satisfy the conditions $2 \leq b \leq c$, and from our reasoning, these are indeed valid.

In the case of $a=3$, our calculations proceed as follows:

$$
\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{6}{4}=\frac{3}{2}, 2(b+1)(c+1)=3 b c, 2 b+2 c+2=b c, b=\frac{2 c+2}{c-2}=\frac{2(c-2)+6}{c-2}=2+\frac{6}{c-2}
$$

Since $c \geq b \geq 3$ and they are integers, the possible values of $c-2$ are: $1,2,3,6$; the corresponding $(b, c)$ pairs are

$$
(8,3),(5,4),(4,5),(3,8),
$$

among which $(4,5)$ and $(3,8)$ also satisfy the size constraints, and these are indeed solutions.

Summarizing: the solutions to the equation are $(2,4,15),(2,5,9),(2,6,7),(3,3,8),(3,4,5)$ and the triples obtained by permuting these (a total of $6+6+6+3+6=27$ solutions).

Based on the work of Szabó 555 Krisztián (Békéscsaba, Széchenyi I. Szki., II. o.t.)