Imagine that the balls have secret identifying numbers. and imagine that we draw out all the balls, whether someone has already won or not.
There are $10!$ ways to draw out the balls, all equally likely.
Player A could win on the first draw. How many sequences correspond to this? The winning red ball could be any of the $3$, and then the rest of the balls could come in any one of $9!$ possible orders, for a total of $(3)(9!)$.
Or else Player A could win on the third draw. That happens if the first ball is black, the second is black, the third is red, and the rest are arranged in any order. The first ball can then be chosen in $7$ ways, and for each choice the second can be chosen in $6$ ways. Then we have $3$ ways for the third, since it has to be red, and then $7!$ for the rest, for a total of $(7)(6)(3)(7!)$.
This should be the second entry in the numerator. It isn't, presumably a typo. No wonder you are puzzled.
Or else Player $A$ could win on the fifth draw. The same reasoning gives $(7)(6)(5)(4)(3)(5!)$ ways. That's correctly written down.
Finally, there could be a win on the $7$-th draw. The number of sequences that give this is correctly written down.
$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.
The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.
$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.
Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.
There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.
One can do a similar calculation for the without replacement case.
For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.
Best Answer
Your answer works. Another way to solve this is to note that because draws are done with replacement, the draws are independent and identically distributed. Thus the random variable $X=\text{number of red balls chosen in 3 draws}$ is a binomial random variable with $n=3$ and $p=\text{probability of red ball on a given draw}=\frac{5}{5+6+8}=\frac{5}{19}$. Thus
$$P(X=1)=\binom{3}{1}\cdot \left(\frac{5}{19}\right)^1\cdot \left(1-\frac{5}{19}\right)^2= \frac{2940}{6859}.$$