Find $a+b+c$, where $a,b,$ and $c$ are the hundreds, tens, and units digits of the six-digit number $123abc$, which is a multiple of $990$.
1. Given the six-digit number \(123abc\) is a multiple of \(990\), we need to find \(a + b + c\). 

2. Since \(990 = 2 \times 3^2 \times 5 \times 11\), the number \(123abc\) must be divisible by \(2\), \(9\), and \(11\).

3. For \(123abc\) to be divisible by \(2\), the units digit \(c\) must be \(0\). Therefore, \(c = 0\).

4. Now, the number \(123ab0\) must be divisible by \(9\). A number is divisible by \(9\) if the sum of its digits is divisible by \(9\). Thus, we have:
   \[
   1 + 2 + 3 + a + b + 0 \equiv 6 + a + b \equiv 0 \pmod{9}
   \]
   This simplifies to:
   \[
   a + b \equiv 3 \pmod{9}
   \]

5. Next, the number \(123ab0\) must be divisible by \(11\). A number is divisible by \(11\) if the alternating sum of its digits is divisible by \(11\). Thus, we have:
   \[
   (1 - 2 + 3 - a + b - 0) \equiv 1 + 3 + b - 2 - a \equiv 2 + b - a \equiv 0 \pmod{11}
   \]
   This simplifies to:
   \[
   b - a \equiv 9 \pmod{11}
   \]

6. We now have two congruences:
   \[
   a + b \equiv 3 \pmod{9}
   \]
   \[
   b - a \equiv 9 \pmod{11}
   \]

7. Solving these congruences, we start with \(b - a \equiv 9 \pmod{11}\). This can be written as:
   \[
   b = a + 9
   \]

8. Substitute \(b = a + 9\) into \(a + b \equiv 3 \pmod{9}\):
   \[
   a + (a + 9) \equiv 3 \pmod{9}
   \]
   \[
   2a + 9 \equiv 3 \pmod{9}
   \]
   \[
   2a \equiv 3 - 9 \pmod{9}
   \]
   \[
   2a \equiv -6 \pmod{9}
   \]
   \[
   2a \equiv 3 \pmod{9}
   \]

9. To solve \(2a \equiv 3 \pmod{9}\), we find the multiplicative inverse of \(2\) modulo \(9\). The inverse of \(2\) modulo \(9\) is \(5\) because \(2 \times 5 \equiv 1 \pmod{9}\). Thus:
   \[
   a \equiv 3 \times 5 \pmod{9}
   \]
   \[
   a \equiv 15 \pmod{9}
   \]
   \[
   a \equiv 6 \pmod{9}
   \]

10. Therefore, \(a = 6\). Substituting \(a = 6\) back into \(b = a + 9\):
    \[
    b = 6 + 9 = 15
    \]

11. Since \(b\) must be a single digit, we need to re-evaluate our steps. We should check for other possible values of \(a\) and \(b\) that satisfy both congruences. 

12. Re-evaluating, we find that \(a = 7\) and \(b = 5\) satisfy both conditions:
    \[
    a + b = 7 + 5 = 12 \equiv 3 \pmod{9}
    \]
    \[
    b - a = 5 - 7 = -2 \equiv 9 \pmod{11}
    \]

13. Therefore, \(a = 7\), \(b = 5\), and \(c = 0\).

14. The sum \(a + b + c = 7 + 5 + 0 = 12\).

The final answer is \(\boxed{12}\).