Solution:

Let $p$ be a prime number. We will prove that the number $A$ of factors of $p$ in $3 \cdot 12 \cdots (9n-6)$ is greater than or equal to the number $B$ of factors of $p$ in $n!$.

We first explain the widely known method for computing $B$. Out of the numbers from $1$ to $n$, exactly $\lfloor n / p \rfloor$ of them are multiples of $p$; they will contribute $\lfloor n / p \rfloor$ "first" factors of $p$ to the product $n!$. In addition, $\left\lfloor n / p^{2} \right\rfloor$ of these numbers are also divisible by $p^{2}$, giving $\left\lfloor n / p^{2} \right\rfloor$ "second" factors of $p$. This continues, and we get
$$B = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^{2}} \right\rfloor + \left\lfloor \frac{n}{p^{3}} \right\rfloor + \cdots,$$
where the sum continues until eventually all of its terms become $0$ due to a lack of terms divisible by very high powers of $p$.

Now we estimate $A$. If $p = 3$, then $A = n$ since every term of the arithmetic sequence $3, 12, 21, \ldots$ is divisible by $3$ but not $9$. In this case it is clear that
$$B = \left\lfloor \frac{n}{3} \right\rfloor + \left\lfloor \frac{n}{3^{2}} \right\rfloor + \left\lfloor \frac{n}{3^{3}} \right\rfloor + \cdots \leq \frac{n}{3} + \frac{n}{3^{2}} + \frac{n}{3^{3}} + \cdots = \frac{n}{2} < n.$$

Now assume that $p \neq 3$. Divide the arithmetic progression $3, 12, 21, \ldots, (9n-6)$ into $\lfloor n / p \rfloor$ blocks of length $p$, discarding any terms that remain; because the common difference $9$ is relatively prime to $p$, each block will have one representative of each congruence class $\bmod\ p$, and in particular exactly one multiple of $p$. Thus the product $3 \cdot 12 \cdot 21 \cdot 30 \cdots (9n-6)$ will have at least $\lfloor n / p \rfloor$ "first" factors of $p$. By the same argument, using blocks of length $p^{2}$, there are at least $\left\lfloor n / p^{2} \right\rfloor$ "second" factors of $p$, and so on, so
$$A \geq \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^{2}} \right\rfloor + \left\lfloor \frac{n}{p^{3}} \right\rfloor + \cdots = B$$
as desired.