1. Determine all triples $(a, b, c)$ where $a, b$, and $c$ are positive integers with $a \leqslant b \leqslant c$ and $a b c=2(a+b+c)$.
1. Suppose $(a, b, c)$ is a solution. From $a \leqslant b \leqslant c$ it follows that $a b c=2(a+b+c) \leqslant 6 c$. Dividing by $c$ gives $a b \leqslant 6$. We see that $a=1$ or $a=2$, because from $a \geqslant 3$ it would follow that $a b \geqslant a^{2} \geqslant 9$.

First, we consider the case $a=2$.

From $a b \leqslant 6$ it follows that $b=2$ or $b=3$. In the first case, from the given $a b c=2(a+b+c)$ it follows that $4 c=8+2 c$ and thus $c=4$. It is easy to verify that the found triplet $(2,2,4)$ is indeed a solution. In the second case, $6 c=10+2 c$, so $c=\frac{5}{2}$. Since $c$ must be an integer, this does not give a solution.

Now we consider the case $a=1$.

We now know that $b c=2(1+b+c)$. Rewriting gives $(b-2)(c-2)=6$. Note that $b-2$ cannot be negative (and thus $c-2$ cannot be negative either). Otherwise, it would have to be that $b=1$, so from $(1-2)(c-2)=6$ it would follow that $c=-4$. But $c$ must be positive.

There are only two ways to write 6 as a product of two non-negative integers, namely $6=1 \times 6$ and $6=2 \times 3$. This gives two possibilities: $b-2=1$ and $c-2=6$, or $b-2=2$ and $c-2=3$. It is easy to verify that the two corresponding triplets $(1,3,8)$ and $(1,4,5)$ are indeed solutions.

The only solutions are thus $(2,2,4),(1,3,8)$ and $(1,4,5)$.