Let's find the three-digit number $\overline{a b c}$, composed of distinct digits, which is equal to the arithmetic mean of the numbers $\overline{b c a}$ and $\overline{c a b}$.
According to the task:

$$
100a + 10b + c = \frac{100b + 10c + a + 100c + 10a + b}{2}
$$

After rearrangement and simplification:

$$
\begin{aligned}
& 7a = 3b + 4c \\
& 7(a - c) = 3(b - c)
\end{aligned}
$$

The left side is divisible by 7, and thus the right side must also be. Since \(b\) and \(c\) are different single-digit numbers, their difference can only be +7 or -7.

For the digits \(b\) and \(c\), the following values are obtained:

$$
\begin{array}{l|llllll}
b & 9 & 8 & 7 & 2 & 1 & 0 \\
c & 2 & 1 & 0 & 9 & 8 & 7 .
\end{array}
$$

By substituting the corresponding values of \(b\) and \(c\) into the equation, we can calculate the appropriate value of \(a\), and thus the following 6 numbers are obtained as the solution to the problem:

$$
\begin{array}{llllll}
592, & 481, \quad 370, & 629, & 518, & 407 .
\end{array}
$$

We can verify the correctness of our solution by substitution.

Fejes László (Makó, József A. g. II. o. t.)

Note: If we do not restrict the difference between the digits, we will get nine additional numbers, all composed of the same digit, as solutions (since if, for example, \(b = c\), the value of \(a\) can also be calculated to be the same from the equation).