3. Give all possible triples $(a, b, c)$ of positive integers with the following properties:

- $\operatorname{gcd}(a, b)=\operatorname{gcd}(a, c)=\operatorname{gcd}(b, c)=1$;
- $a$ is a divisor of $a+b+c$;
- $b$ is a divisor of $a+b+c$;
- $c$ is a divisor of $a+b+c$.

(For given numbers $x$ and $y$, $\operatorname{gcd}(x, y)$ denotes the greatest common divisor of $x$ and $y$.)
3. Note that the problem is symmetric in $a, b$, and $c$. This means that if we swap $a$ and $b$, or $a$ and $c$, or $b$ and $c$ everywhere, we get the same conditions. Therefore, we first look at the solutions with $a \leqslant b \leqslant c$. The other solutions can then be found by swapping the values of $a$, $b$, and $c$ among themselves.

Since $a$ and $b$ are positive, it follows that $a+b+c > c$. Since $c$ is the largest of the three, it also holds that $a+b+c \leqslant 3c$. The given condition that $a+b+c$ must be a multiple of $c$ now leads to two possibilities: $a+b+c=2c$ or $a+b+c=3c$. We will examine these two cases separately.

If $a+b+c=3c$, then $a$, $b$, and $c$ must all be equal, because $c$ is the largest of the three and $3c$ would otherwise be greater than $a+b+c$. The greatest common divisor $\operatorname{gcd}(b, c)$ is thus equal to $\operatorname{gcd}(c, c)=c$ and this must be 1 according to the conditions. We thus find that $(a, b, c)=(1,1,1)$ and we see that this is indeed a solution because $\operatorname{gcd}(1,1)=1$ and 1 is a divisor of $1+1+1$.

If $a+b+c=2c$, then we have that $c=a+b$. We know that $b$ must be a divisor of $a+b+c=2a+2b$. Since $a>0$, it holds that $2a+2b>2b$. Since $b \geqslant a$, it holds that $2a+2b \leqslant 4b$. Since $2a+2b$ must be a multiple of $b$, we thus have only two possibilities: $2a+2b=3b$ and $2a+2b=4b$. We will examine these two cases again separately.

If $2a+2b=3b$, then we find $b=2a$. Just like in the first case, we now find that $a=\operatorname{gcd}(a, 2a)=\operatorname{gcd}(a, b)=1$. We thus find $b=2$ and $c=a+b=3$. The found triplet $(a, b, c)=(1,2,3)$ is indeed a solution because $\operatorname{gcd}(1,2)=\operatorname{gcd}(1,3)=\operatorname{gcd}(2,3)=1$ and $1+2+3=6$ is divisible by 1, 2, and 3.

If $2a+2b=4b$, then we find $a=b$. Again, we see that $a=\operatorname{gcd}(a, a)=\operatorname{gcd}(a, b)=1$. From $b=a=1$ it follows that $c=a+b=2$ and thus $(a, b, c)=(1,1,2)$. This is indeed a solution because $\operatorname{gcd}(1,1)=\operatorname{gcd}(1,2)=1$ and $1+1+2=4$ is divisible by 1 and 2.

The solutions with $a \leqslant b \leqslant c$ are thus: $(a, b, c)=(1,1,1)$, $(a, b, c)=(1,1,2)$, and $(a, b, c)=(1,2,3)$. This gives a total of ten solutions $(a, b, c)$, namely

$$
(1,1,1), \quad(1,1,2),(1,2,1),(2,1,1), \quad(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)
$$