The following cards are dealt with $3$ people at random so that every one of them gets the same number of cards: $$R_1, R_2, R_3, B_1, B_2, B_3, Y_1, Y_2, Y_3$$ where $R$, $B$ and $Y$ denote red, blue and yellow, respectively. Find the probability that everyone gets a red card.
A total number of ways we can distribute $9$ cards to $3$ people so that each gets an equal number of cards is equal to
$$\binom{9}{3} \times \binom{6}{3}$$
Next, we can give a red card to each of them in $3!$ ways. Then divide the remaining $6$ cards in $\binom{6}{2} \times \binom{4}{2}$ ways. Hence, the required probability is
$$ \frac{3! \times \binom{6}{2} \times \binom{4}{2}}{\binom{9}{3} \times \binom{6}{3}} $$
Is my argument correct?
Your solution is correct, but since only probability has been asked for, it is much easier way to compute it directly by just considering where the red cards need to go, since where others go doesn't matter.
The first red card can go anywhere, thus $Pr = \dfrac68\cdot\dfrac37$