Other answers have explained how to work the problem correctly, but not what’s wrong with your solution. You correctly calculated that there are $10$ distinguishable combinations of colors, all of which are possible in every version of the problem, and observed that only one of them contains one ball of each color. If the $10$ different combinations were equally likely to be drawn, your answer would be correct. However, they are not equally likely to be drawn.
This is probably easiest to see in the first version, in which we have specific numbers of balls of each color. Imagine that the nine balls have invisible labels: the red balls are labelled $1,2$, and $3$, the green balls are labelled $4,5$, and $6$, and the blue balls are labelled $7,8$, and $9$. You reach into the box and without looking grab three balls. If the numbers were suddenly made visible, you’d find that you’d chosen a $3$-element subset of $\{1,2,3,4,5,6,7,8,9\}$. Each of these subsets is equally likely to be chosen. Thus, the probability of getting one of your $10$ color combinations can be found by dividing the number of subsets representing that combination by the total number of subsets, which is $\binom93=84$.
The all-red combination, for instance, is obtained only when you draw the set $\{1,2,3\}$, so the probablity of getting this color combination is only $\frac1{84}$. If all $10$ of the possible combinations were equally likely, the probabilities of getting them would add up to only $10\cdot\frac1{84}=\frac5{42}$, which is absurd. Thus, they cannot be equally likely, and this fact immediately invalidates your reasoning.
You can of course go on to use this approach to calculate the correct probability. In order to get one ball of each color, you must get a set containing exactly one member of $\{1,2,3\}$, one member of $\{4,5,6\}$, and one member of $\{7,8,9\}$. There are $3^3=27$ such sets, so the probability of getting balls of all three colors is $\frac{27}{84}=\frac9{28}$.
Similarly, you can calculate that the probability of getting two red balls and one green ball is $\frac9{84}=\frac3{42}$, since there are $\binom32\binom31=9$ $3$-elements subsets of $\{1,2,3,4,5,6,7,8,9\}$ that result in that color combination. Each of the other two-color combinations of course has the same probability.
This is enough to show that you can’t safely assume that the different combinations of colors are equally likely; we don’t have to demonstrate it anew for each of the other versions of the problem.
Best Answer
The problem, with that hypergeometric distribution way of solving it, is that ${8}\choose{1}$. That would suggest, that there are 5+4+4+8 = 21 balls in total. You could for example distinguish between the cases for the color of the last ball: ${5\choose 3} \cdot {4\choose 2} \cdot {4 \choose 1} + {5\choose 2} \cdot {4\choose 3} \cdot {4 \choose 1} + {5\choose 2} \cdot {4\choose 2} \cdot {4 \choose 2}$