338. Find all such triples of pairwise coprime natural numbers such that the sum of any two of them is divisible by the third.
$\triangle$ Let's denote these numbers by $a, b,$ and $c$. Then

$$
(a+b): c, \quad(a+c): b, \quad(b+c): a
$$

We will apply the same method of solution as in problem 335. Since the numbers $a, b,$ and $c$ are pairwise coprime, the system of three divisibilities is equivalent to one:

$$
(a+b+c): a b c
$$

(check this!). Let $a \leqslant b \leqslant c$. We get:

$$
a b c \leqslant a+b+c \leqslant c+c+c=3 c
$$

from which $a b c \leqslant 3 c, a b \leqslant 3$. We will consider three cases.

1) Suppose $a=b=1$.

Then the divisibility $(a+b): c$ takes the form $2: c$, so $c=1$ or $c=2$. In each of these subcases, the divisibilities $(a+c): b$ and $(b+c): a$ are satisfied.

2) Suppose $a=1, b=2$.

Then $3: c$, so $c=1$ or $c=3$. The value $c=1$ should be discarded, and for $c=3$ we get the valid divisibilities $4: 2$ and $5: 1$.

The case $a=2, b=1$ does not need to be considered since the inequality $a \leqslant b$ must hold.

3) Suppose $a=1, b=3$.

Then $4: c$, so $c=4$. But in this case, the divisibility $(a+c): b$ takes the form $5: 3$, which is false.

Answer: $1,1,1 ; 1,1,2 ; 1,2,3$.