The following analysis is correct in spirit but wrong in detail, since I misread the problem slightly; the correct analysis is at the bottom.
It’s true that balls of the same color are indistinguishable, but the sequence of colors does matter. If you had at least $3$ balls of each color, you could get every possible sequence of the three colors: $BBB,BBG,RBB,GRG,GGR,GRB$, etc., and there are $3^3=27$ such sequences. However, you have only one blue ball and two green balls, so some of these sequences are impossible. To get the final result you can either count the possible sequences directly or count the impossible sequences and subtract from $27$. It’s not hard to count the possible sequences directly.
There is one way to have three red balls, $RRR$, and it’s impossible to have three green or three blue balls. Thus, there’s just one way to have all three balls the same color.
Now let’s use exactly two colors. There are three ways to have two red balls and one green ball: $RRG,RGR,GRR$. Similarly, there are three ways to have two red balls and one blue ball, three ways to have two green balls and one red ball, and three ways to have two green balls and one blue ball. We can’t have two blue balls, so we’ve covered the two-color possiblities: there are $12$ of them.
Finally, if we use all three colors, we have one ball of each color, and there are $3!=6$ different arrangements. I get a total of $1+12+6=19$ possible sequences.
As a check, let’s make sure that there are $27-19=8$ impossible sequences. There are two one-color sequences that are impossible, $BBB$ and $GGG$. The only other impossible sequences are those requiring two blue balls. There are three that use a green ball with the two blue balls: $BBG,BGB,GBB$. There are another three that replace the green ball with a red ball. Thus, we do indeed have a total of $2+3+3=8$ impossible sequences.
Added: As Henry points out, I somehow misread the problem. Since there are in fact three blue balls, the only impossible sequence of colors is $GGG$, and the other $27-1=26$ sequences are all possible. I’ll leave the incorrect analysis, since it may be useful to someone attempting a similar problem.