Using the digits a, b, and c, we construct the three-digit number $abc$ and the two-digit numbers $ab$, $bc$, and $ca$. Find all possible values of $a$, $b$, and $c$ such that $\frac{abc + a + b + c}{ab + bc + ca}$ is an integer.

Hint: Show that the denominator is always divisible by 11 and that $a + c = 11$.

Helpful Facts: $abc = 100a + 10b + c$, $ab = 10a + b$.
Let's observe that

$$
a b+b c+c a=(10 a+b)+(10 b+c)+(10 c+a)=11(a+b+c)
$$

so the denominator of the fraction is divisible by 11. Since the fraction is an integer, the numerator

$$
a b c+a+b+c=(100 a+10 b+c)+a+b+c=101 a+11 b+2 c
$$

is also divisible by 11. Since

$$
101 a+11 b+2 c=11(9 a+b)+2(a+c)
$$

it follows that $a+c$ is divisible by 11. Since $a$ and $c$ are digits and $a \neq 0$, $1 \leqslant a+c \leqslant 18$, hence $a+c=11$. Substituting $c$ in the given expression, we get

$$
\begin{aligned}
\frac{a b c+a+b+c}{a b+b c+c a} & =\frac{11(9 a+b+2)}{11(b+11)} \\
& =\frac{9 a+b+2}{b+11} \\
& =\frac{b+11+9 a-9}{b+11} \\
& =1+\frac{9(a-1)}{b+11}
\end{aligned}
$$

and $\frac{9(a-1)}{b+11}$ is an integer. The digit $a$ cannot be 1, because $a+c=11$. We observe that if $b+11$ is not divisible by 3, we would have that $b+11$ divides $a-1$, which is impossible since $b+11 > a-1$. Thus, $b+11$ is equal to 12, 15, or 18. So $b=1, 4$, or 7.

- If $b=1$, since $\frac{9(a-1)}{12}=\frac{3(a-1)}{4}$ is an integer, we have $a=5$ or 9, generating the numbers 516 and 912.
- If $b=4$, $\frac{9(a-1)}{15}=\frac{3(a-1)}{5}$ is an integer, so $a=6$, generating the number 645.
- If $b=7$, $\frac{9(a-1)}{18}=\frac{a-1}{2}$, so $a=3, 5, 7$, or 9, generating the numbers 378, 576, 775, and 972.