Let's determine the distinct digits $A, B, C$, if in the decimal system

$$
\overrightarrow{A B C}=\overline{A B} \cdot C+\overline{B C} \cdot A+\overline{C A} \cdot B
$$
1. Let's write the two- and three-digit numbers in sum form - taking into account the place value of each digit - and then rearrange the right side of the equation:

$$
\begin{gathered}
100 A+10 B+C=(10 A+B) \cdot C+(10 B+C) \cdot A+(10 C+A) \cdot B= \\
=11(A B+B C+A C)
\end{gathered}
$$

Accordingly, both sides are multiples of 11. The coefficients on the left side differ from a multiple of 11 by only 1, thus

$$
100 A+10 B+C=11(9 A+B)+(A-B+C)
$$

Therefore, $(A-B+C)$ is also divisible by 11.

Each of the digits in question appears as a leading digit, so:

$$
1 \leq A, B, C \leq 9
$$

Moreover, they are distinct integers. Therefore, on one hand,

$$
A+C-B \leq 9+8-1=16
$$

and on the other hand,

$$
A+C-B \geq 1+2-9=-6
$$

Between these two limits, only two numbers are divisible by 11, namely 0 and 11. We will examine these two cases separately.

2. If $A-B+C=0$, substitute $B$ with $A+C$ in (1). Dividing immediately by 11:

$$
10 A+C=(A+C)^{2}+A C
$$

or

$$
10 A-A^{2}-3 A C=A(10-A-3 C)=C^{2}-C=C(C-1)
$$

According to this, the right side is divisible by $A$, let's write it as: $C(C-1)=K A$, where $K \geq 0$ is an integer.

The case $K=0$ immediately gives a solution, from which $C=1$, and from the equation

$$
A(7-A)=0
$$

we get $A=7$, then $B=8$, and indeed it holds that

$$
78 \cdot 1+81 \cdot 7+17 \cdot 8=781
$$

However, for $K>0$ there is no solution, because in this case on one hand $C \neq 1, C \geq 2$ and on the other hand from the equation divided by $A$,

$$
10-A-3 C=K \geq 1
$$

thus $C<3, C \leq 2$. But for $C=2$, (3) becomes $A^{2}-4 A+2=0$, and this has no integer solution.

3. If $A-B+C=11$, again by eliminating $B=A+C-11$, similar steps from (1) yield

$$
10 A+C-10=(A+C-11)(A+C)+A C
$$

which we rearrange as follows:

$$
3(7 A+4 C-A C-4)=A^{2}+C^{2}-2
$$

Due to the left side, the right side is also divisible by 3, or when divided by 3, the remainder of $\left(A^{2}+C^{2}\right)$ is 2. Since the squares of numbers in the form $3 k, 3 k+1, 3 k-1$ are respectively $3 m$, $(3 m+1)$, and $(3 m+1)$, neither $A$ nor $C$ is divisible by 3. Therefore, $A+C \leq 8+7=15$, and on the other hand, since $B \geq 1$, $A+C \geq 12$. Now we examine the values 12, 13, 14, and 15 for $A+C$. For this, we rearrange (4) as follows:

$$
3(A-4)(7-C)=A^{2}+C^{2}-74
$$

For the sum $A+C=12$, it can be expressed as $8+4$ and $7+5$ - in both orders - and the value of the right side is 6 and 0, respectively, in both cases

$$
(A-4)(7-C)=2, \quad \text { or } \quad 0
$$

If the right side is 2, then $A \neq 4$, but for $A=8$ and $C=4$, the equation does not hold. The other possibility gives $C=7$, and $A=5, B=1$ as a solution, satisfying the requirement:

$$
517=51 \cdot 7+17 \cdot 5+75 \cdot 1
$$

The other three cases do not provide a solution. For $A+C=14$, the smaller number is 6 due to the limit and $A \neq C$, but this is a multiple of 3. For $A+C=15$, only $8+7$ is possible, and for $A+C=13$, only $8+5$ is possible, but with these, the right side is $39=3 \cdot 13$, and $15=3 \cdot 5$, respectively, while the left side cannot have a factor of 13 or 5.

We have considered all possibilities and found the following two digit-triplets to be suitable:

$$
A=7, B=8, C=1 \quad \text { and } \quad A=5, B=1, C=7 \text {. }
$$

Remarks. 1. The rearrangement in (2) is not just a makeshift trick but can be used for any number of digits, and it is a useful criterion for divisibility by 11. For any even exponent, $10^{2 k}-1$ is divisible by $10^{2}-1=11 \cdot 9$ and for any odd exponent, $10^{2 k+1}+1$ is divisible by $(10+1)$. In other words: the remainder of $10^{n}$ when divided by 11 is +1 or $(-1)$ depending on whether $n$ is odd or even. Therefore, instead of the number $N=a_{n} a_{n-1} \ldots a_{1} a_{0}$, it is sufficient to examine the difference $J=\left(a_{0}+a_{2}+a_{4}+\ldots\right)-\left(a_{1}+a_{3}+a_{5}+\ldots\right)$ formed from its digits. $N$ is divisible by 11 if and only if $J$ is divisible by it.

2. We did not evaluate those solutions that obtained the suitable digit-triplets based on computer calculations (see the September issue of our journal for the competition announcement).