What is my mistake in this counting problem? 
Question: How many 6-digit numbers, written in decimal notation, have at least one 1, one 2, and one 3 among their digits?

Define $A, B, C$ to be the set of $6$-digit numbers with at least one $1$, at least one $2$, and at least one $3$, respectively.
We may count the cardinality of $A$ with two cases: either $A$ has first digit $1$ or it is not a first digit. If the first digit is $1$, then there are $1$ choices for the first digit ($1$) and $10^5$ choices for the remaining digits. If the first digit of $A$ is not $1$, then there are $8$ choices ($\{2,3,\dots,9 \}$) choices for the first digit, $5$ choices for where $1$ is, and $10^4$ choices for the remaining digits. In total, there are $8\cdot5\cdot10^4 + 10^5$ digits with at least one $1$. By symmetry, we see that $|A| = |B| = |C|$.
We shall now count $A\cap B$ in a similar manner. Either the first digit is a $1$ or a $2$, or it isn't. There are $2$ choices for the first digit ($1$ or $2$), and $5$ choices for the other digit which wasn't first. Then there are $10^4$ choices for the remaining digits. If the first digit is not a $1$ or a $2$, then there are $7$ choices for the first digit ($\{3,4,\dots,9\}$) and $5\cdot4$ choices for the $1$ and $2$, and $10^3$ choices for the remaining digits. By symmetry, we have that $|A \cap B| = |A \cap C| = |B \cap C| = 2\cdot 5 \cdot 10^4 + 7\cdot 5 \cdot 4 \cdot 10^3$.
Finally, we count $A \cap B \cap C$ using the same method. Either the first digit is a $1, 2$ or $3$ or it is not. If it is, then we have $3$ choices for which of the three it is, then $5\cdot4$ choices for where to place the other two. Then we have $10^3$ ways to fill the remaining digits. If the first digit is not a $1,2$ or $3$, then we have $6$ choices for the first digit, and $5\cdot4\cdot3$ ways to pick places for the $1,2,3$, and $10^2$ ways to fill the remaining digits.
By the Principle of Inclusion-Exclusion, we have that $$ |A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|, $$ so we have that $$ 3(8\cdot5 \cdot 10^4 + 10^5) - 3(2 \cdot 5 \cdot 10^4 + 7\cdot 5 \cdot 4 \cdot 10^3) + 3 \cdot 5 \cdot 4 \cdot 10^3 + 6 \cdot 5 \cdot 4 \cdot 3 \cdot 10^2, $$ which is much larger than the supposed answer that you would get using complementary counting.

We have to find $|A \cap B \cap C|$.
Note that the complement of $A$ is the set of $6$-digit numbers which have no digit $1$, therefore $|A^c|=8\cdot 9^5$. Hence in your attempt you should have $|A|=9\cdot 10^5-8\cdot 9^5$ which less than $8\cdot5\cdot10^4 + 10^5$ because you are overcounting! For instance the number $211111$ is counted several times in $8\cdot5\cdot10^4$.
Similarly, $|B^c|=|C^c|=8\cdot 9^5$,  $|A^c \cap B^c|=|B^c \cap C^c|=|A^c \cap C^c|=7\cdot 8^5$ and $|A^c \cap B^c \cap C^c|=6\cdot 7^5$.
Finally, by the Principle of Inclusion-Exclusion, we have
$$\begin{align*}|A \cap B \cap C|&=9\cdot 10^5-|A^c \cup B^c \cup C^c|\\
&=9\cdot 10^5-|A^c| - |B^c| - |C^c|+(|A^c \cap B^c| + |A^c \cap C^c| + |B^c \cap C^c|)\\
&\qquad - |A^c \cap B^c \cap C^c|\\
&=9\cdot 10^5-3\cdot 8\cdot 9^5+3\cdot 7\cdot 8^5-6\cdot 7^5=70110.
\end{align*}$$
