Hint: One approach is to define four functions. Let $OE(n)$ be the number of words of length $n$ which have an odd number of $1$'s and an even number of $2$'s and three similar ones. Write the recurrence relations between these four. They stay closely balanced, so they are all just about $\frac {3^n}4$ in each one. Maybe you can find a proof of the small correction (note that this is not an integer, which the final result must be)
Added: you are right that $EO(n)=OE(n)$ by symmetry. That is a good thing to think about, as it gets you down to three functions. You need to write a recurrence for each function separately. A word where both are odd can come from a word where both are odd by adding a 3, from EO by adding a 1, or from OE by adding a 2. So $OO(n)=EO(n-1)+OE(n-1)+OO(n-1)$ Your starting condition is $EE(0)=1, \text{others}(0)=0$ because the empty word is even in both. I made a spreadsheet to calculate the first dozen values or so. You can then see a pattern in the corrections, which you can prove from the recurrences. Your final answer is then $EO(n)$
This can be done recursively. We have $a_1 = 3$ and $a_2 = 7$ by inspection.
Then for $n>2$, we see that there are three possible ways to start a sequence: $AB$, $B$, or $C$. Depending on which one we choose, we recurse into a smaller sequence of known size, where the counting process is repeated all over again.
This suggests the recurrence $a_n = 2a_{n-1} + a_{n-2}$ with $a_1 = 3$ and $a_2 = 7$.
To get a closed form, note that the characteristic polynomial is $x^2 - 2x - 1 = 0$, or $(x-1)^2-2 = 0$, which has roots $1 + \sqrt{2}$ and $1 - \sqrt{2}$. Therefore:
$a_n = A(1 + \sqrt{2})^n+B(1 - \sqrt{2})^n$ for some unknown $A,B$.
But we already know $a_1$ and $a_2$:
$3 = A(1 + \sqrt{2})^1+B(1 - \sqrt{2})^1$
$7 = A(1 + \sqrt{2})^2+B(1 - \sqrt{2})^2$
Solving this system yields $A = \frac{1}{2}+\frac{1}{\sqrt{2}}, B = \frac{1}{2}-\frac{1}{\sqrt{2}}$. Therefore:
$$a_n = (\frac{1}{2}+\frac{1}{\sqrt{2}})(1 + \sqrt{2})^n+(\frac{1}{2}-\frac{1}{\sqrt{2}})(1 - \sqrt{2})^n$$
Simplifying:
$$a_n = \frac{1}{2} ((1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1})$$
Best Answer
Hint: Let $a_n$ be the number of strings which satisfy the conditions of length $n$.
If a string begins with $b$ or $c$, then you are left with a string of length $n-1$, which can start with any character. How many strings are there of this type (it is related to $a_{n-1}$)?
If a string begins with $a$, then the second letter must be a $b$ or $c$, then you are left with a string of length $n-2$, which can start with any character. How many strings are there of this type (it is related to $a_{n-2}$)?