For an integer $n \ge 2$, let us consider the number of ways of filling the $n$ boxes lined up from left to right by putting $n$ balls one-by-one into the boxes starting from the left-most one, assuming that any of the balls can have any of the three colors. Denote by $a_n$ the possible number of ways of filling the boxes so as to satisfy the condition of the problem, starting with a red ball going into the left-most box. We see that the number of ways will be the same $a_n$ if we start with a blue ball instead of a red one, and this is true also if we start with a yellow one.

Next, suppose $n \ge 4$, and we start with a red ball, and count how many red balls were used before the first non-red ball is put into a box. Call this number $k$. Then by the assumption of the problem, $k \ge 2$. Now consider the following two cases:

(1) When $k=2$: In this case, either a blue or a yellow ball has to go into the third box, and in either case the number of ways of filling the remaining $n-2$ boxes so as to satisfy the condition of the problem is $a_{n-2}$, and therefore there are $2a_{n-2}$ ways of filling the $n$ boxes so as to satisfy the condition.

(2) When $k \ge 3$: In this case, if we remove the left-most box (containing a red ball), the left-most box for the remaining $n-1$ boxes has a red ball and the number of ways of filling the remaining $n-2$ boxes must be $a_{n-1}$ in order to satisfy the condition.

Thus we get the recurrence formula: $a_n = a_{n-1} + 2a_{n-2}$.

When $n=2,3$ the only way to fill boxes starting with a red ball and satisfying the condition is to use red balls only, so we have $a_2 = 1$ and $a_3 = 1$.

If we use the recurrence formula above repeatedly, we obtain $a_{12} = 683$, and therefore, the desired answer is $3 \times 683 = 2049$.