A 7 is a divisor of the number $\overline{A B C C B A}=100001 A+10010 B+1100 C$ and the number $100 A+10 B+C$, where $A, B, C$ are different digits, and $A \neq 0$.

Since

$$
\overline{A B C C B A}=1000 \overline{A B C}+(A+10 B+100 C)
$$

it follows that $\overline{C B A}=100 C+10 B+A$ must also be divisible by 7. Therefore, 7 must be a divisor of $\overline{A B C}-\overline{C B A}=99 A-99 C=99(A-C)$. Since 99 is not divisible by 7, and 7 is a prime number, $(A-C)$ must be divisible by 7.

This implies the following possibilities:
a) $A=1, C=8$.

The conditions $7 \mid \overline{1 B 8}$ and $7 \mid \overline{8 B 1}$ are only satisfied by $B=6$, yielding the number: 168861.
b) $A=2, C=9$.

The conditions $7 \mid \overline{2 B 9}$ and $7 \mid \overline{9 B 2}$ are only satisfied by $B=5$, yielding the number: 259952.
c) $A=8, C=1$.

The conditions $7 \mid \overline{8 B 1}$ and $7 \mid \overline{1 B 8}$ are only satisfied by $B=6$, yielding the number: 861168.
d) $A=9, C=2$.

The conditions $7 \mid \overline{9 B 2}$ and $7 \mid \overline{2 B 9}$ are only satisfied by $B=5$, yielding the number: 952259.
e) $A=7, C=0$.

The condition $7 \mid \overline{7 B 0}$, but $B \neq 0, B \neq 7$, because $A, B, C$ are different. There is no solution here.

The four six-digit numbers above satisfy the conditions of the problem.

Based on the work of Zsuzsanna Sevecsek (Tatabánya, Árpád Gymnasium, 10th grade).
Find all 7-divisible six-digit numbers of the form $\overline{A B C C B A}$, where $\overline{A B C}$ is also divisible by 7. ( $A$, $B, C$ represent different digits.)