When dealing without replacement, the result is just
$$\frac{{5 \choose 1}\cdot{6 \choose 1}\cdot{8 \choose 1}}{19 \choose 3}\approx0.2477$$
The denominator $${19 \choose 3}=\frac{19!}{\color{blue}{3!}\cdot16!}$$ serves to account for different orderings so you don't need to multiply by $3!$
Note that your result is greater than one and a probability must be between $0$ and $1$.
When working with replacement, this becomes a multinomial distribution. Letting $X_1$, $X_2$, and $X_3$ denote the number of red, blue, and green balls selected, respectively, we have
$$\mathsf P(X_1=X_2=X_3=1)=\frac{3!}{1!\cdot1!\cdot1!}\cdot\frac{5}{19}\cdot\frac{6}{19}\cdot\frac{8}{19}\approx0.2099$$
which agrees with your result.
R Simulation Without Replacement:
> urn = c(rep("red",5),rep("blue",6),rep("green",8))
> u = replicate(10^6, length(unique(sample(urn,3,repl=F))))
> mean(u == 3)
[1] 0.247378
R Simulation With Replacement:
> urn = c(rep("red",5),rep("blue",6),rep("green",8))
> u = replicate(10^6, length(unique(sample(urn,3,repl=T))))
> mean(u == 3)
[1] 0.210267
There will be an urn in which the number $r$ of red balls will satisfy
$r\in\left\{ 0,1,2\right\} $.
To be found is an answer to the question: "for which of the options $r=0,1,2$ will $P(S)$ be maximal?"
Let $E$ denote the event that this
urn will be chosen at random and let $S$ denote the event that two
balls are chosen that have the same color.
If $r=0$ then $S=E$ hence $P\left(S\right)=P\left(E\right)=\frac{1}{2}$.
If $r\in\left\{ 1,2\right\} $ then $$P\left(S\right)=P\left(E\right)P\left(S\mid E\right)+P\left(E^{\complement}\right)P\left(S\mid E^{\complement}\right)=\frac{1}{2}\left[P\left(S\mid E\right)+P\left(S\mid E^{\complement}\right)\right]$$
Here $$P\left(S\mid E\right)=\frac{r}{5}\frac{5-r}{5}+\frac{5-r}{5}\frac{4-r}{4}$$
and $$P\left(S\mid E^{\complement}\right)=\frac{5-r}{5}\frac{r}{5}+\frac{r}{5}\frac{r-1}{4}$$
(do you see why?).
Now substitute $r=1,2$ and draw conclusions.
Best Answer
You need to multiply your answer by $6$.
Your probability is for a red ball followed by a green ball followed by a blue ball. The probability of a green ball followed by a blue ball followed by a red ball would be $\displaystyle \frac{7}{21} \cdot \frac{9}{20} \cdot \frac{5}{19}$ which is also $ \dfrac {3}{76}$ and altogether there are $3!=6$ possible orderings.
This would give $\dfrac{9}{38}\approx 0.2368$ as the probability of three different colours in any order.