Let's determine those $A B C$ numbers which are equal to the sum of the numbers $A B, A C, B A, B C, C A, C B$. $(A, B, C$ are digits, each number is understood in the decimal system.)
I. solution. According to the condition, the sum of the two-digit numbers formed from these pairs is equal to the three-digit number $100 A+10 B+C$, that is,

$$
22(A+B+C)=100 A+10 B+C
$$

Rearranging and dividing by 3, we get $26 A=4 B+7 C$. If we substitute the largest possible values, 9, for $B$ and $C$, the value of the expression on the right side is not greater than 99, from which it follows that $A \leqq 3$. We can also read from the equation that $C$ is even.

Let's write $2 D$ for $C$, where $D \leqq 4$, so

$$
13 A=2 B+7 D
$$

Let's examine the cases $A=1,2,3$ separately. If $A=1$, then

$$
13=2 B+7 D
$$

from which $D=1, C=2$, and thus $B=3$, so the three-digit number is 132.

If $A=2$, then $26=2 B+7 D$, from which $D$ is even, so $D=2$ is the only possibility, then $C=4, B=6$, the three-digit number is 264.

Finally, in the case of $A=3$, $39=2 B+7 D$. Now $D$ can only be odd and $D \leqq 4$. But $D=1$ is not possible, because then $B>9$, so $D=3$, and then $B=9, C=6$.

So the three-digit number is: 396.

In summary, we have found three numbers that satisfy the requirement, and these are: 132, 264, 396.

II. solution. It can be seen from equation (1) that the three-digit number is divisible by 22. Subtract $(A+B+C)$ from both sides of (1)

$$
21(A+B+C)=9(11 A+B)
$$

Dividing by 3,

$$
7(A+B+C)=3(11 A+B)
$$

From this, $(A+B+C)$ is divisible by 3, but if the sum of the digits of the three-digit number is divisible by 3, then the number itself is also, and thus $ABC$ is divisible by 66. We already saw in the first solution that $A \leqq 3$, so we only need to consider the next multiples of 66: $132,198,264,330$ and 396.

It is easy to check that 198 and 330 do not meet the requirements.

III. solution. The equation (2) in the first solution can be rearranged as follows:

$$
13(A-D)=2(B-3 D) . \quad(D \leqq 4)
$$

Thus $(B-3 D)$ is divisible by $13$, but $0 \leqq B \leqq 9$ and $0 \leqq D \leqq 4$ so

$$
-12 \leqq B-3 D \leqq 9
$$

so $B=3 D$, and thus $A=D$, that is, the digits of the solution are of the form $A, 3 A$ and $2 A$. For $A=1,2,3$, the resulting three-digit numbers are: 132, 264, and 396.