Given the problem, we need to divide the set $\{1, 2, 3, \ldots, 63\}$ into three non-empty disjoint sets $A$, $B$, and $C$. Let the product of the numbers in these sets be $a$, $b$, and $c$, respectively. We aim to maximize the greatest common divisor (GCD) of these three products, $\gcd(a, b, c)$.

First, calculate the product of all numbers from 1 to 63:

$$P = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot 63 = 63!$$

Since $A$, $B$, and $C$ together contain each of the numbers exactly once, their combined product is also $63!$. Therefore:

$$a \cdot b \cdot c = 63!$$

To find $\gcd(a, b, c)$, we consider the prime factorizations. We utilize the principle that the GCD is maximized when the prime factors are evenly distributed among $a$, $b$, and $c$.

Calculate the prime factorization of $63!$. For a prime $p$, the exponent of $p$ in $63!$ is given by:

$$e_p = \left\lfloor \frac{63}{p} \right\rfloor + \left\lfloor \frac{63}{p^2} \right\rfloor + \left\lfloor \frac{63}{p^3} \right\rfloor + \cdots$$

Compute for each prime number up to 63:

- Prime 2:
$$e_2 = \left\lfloor \frac{63}{2} \right\rfloor + \left\lfloor \frac{63}{4} \right\rfloor + \left\lfloor \frac{63}{8} \right\rfloor + \left\lfloor \frac{63}{16} \right\rfloor + \left\lfloor \frac{63}{32} \right\rfloor = 31 + 15 + 7 + 3 + 1 = 57$$

- Prime 3:
$$e_3 = \left\lfloor \frac{63}{3} \right\rfloor + \left\lfloor \frac{63}{9} \right\rfloor + \left\lfloor \frac{63}{27} \right\rfloor = 21 + 7 + 2 = 30$$

- Prime 5:
$$e_5 = \left\lfloor \frac{63}{5} \right\rfloor + \left\lfloor \frac{63}{25} \right\rfloor = 12 + 2 = 14$$

- Prime 7:
$$e_7 = \left\lfloor \frac{63}{7} \right\rfloor + \left\lfloor \frac{63}{49} \right\rfloor = 9 + 1 = 10$$

- Higher Primes:
Simply calculate based on the limited number of occurrences up to 63.

Distribute these exponents evenly among $a$, $b$, and $c$ to maximize the GCD. Note that if a certain power cannot be distributed evenly, a small remainder may be distributed among one or two products, minimizing impact on the GCD:

- Exponents Distribution:
- 2: Divide $57$ into $19 + 19 + 19$
- 3: Divide $30$ into $10 + 10 + 10$
- 5: Divide $14$ into $4 + 5 + 5$ or another combination maximizing GCD
- Continue similarly for all smaller primes up to 63.

The resulting maximum GCD, evenly distributing prime factors, would be:

$$\boxed{2^{19} \cdot 3^{10} \cdot 5^4 \cdot 7^3 \cdot 11 \cdot 13 \cdot 17 \cdot 19}$$

Where the excess factors are allocated optimally for maximizing the GCD.