First we describe the sample space. This consists of all sequences of length $6$ made up of letters chosen from our $3$-letter alphabet. There are $3^6$ such sequences, and by assumption they are all equally likely.
Now we have to count the "favourables," the sequences of length $6$ that have $3$ A's, $2$ B's, and (therefore) $1$ C.
The location of the A's can be chosen in $\dbinom{6}{3}$ ways. For each such way, there are $\dbinom{3}{2}$ ways to choose the location of the B's, for a total of $\dbinom{6}{3}\dbinom{3}{2}$.
Now divide by $3^6$.
Remark: Let us change the problem so that we still have independence, but we drop the requirement of equally likely. So suppose that A has probability $a$, B has probability $b$, and C has probability $c$, where $a+b+c=1$.
Any particular sequence made up of $3$ A's, $2$ B's, and $1$ C has probability $a^3b^2c^1$. There are $\dbinom{6}{3}\dbinom{3}{2}$ such sequences, so our probability is $\binom{6}{3}\binom{3}{2}a^3b^2c^1.$ In the case of your question, we have $a=b=c=\dfrac{1}{3}$.
The same basic strategy can be used for any similar problem. For example, if we modify your problem slightly to have $4$ equally likely outcomes A, B, C, D, then the probability of $4$ A's, $2$ B's, $1$ C and $3$ D's in $10$ trials is $\dbinom{10}{4}\dbinom{6}{2}\dbinom{4}{1}$ divided by $4^{10}$.