1. Define the circles and their properties:
- Let $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ be three pairwise externally tangent circles with radii $1$, $2$, and $3$, respectively.
- Denote the centers of $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ by $C$, $A$, and $B$, respectively.
- Let $\Omega$ be the circle that passes through the centers of $\Gamma_2$ and $\Gamma_3$ and is externally tangent to $\Gamma_1$ at a point $P$.
- Denote the center of $\Omega$ by $O$ and its radius by $r$.

2. Collinearity and distances:
- Since $\Gamma_1$ and $\Omega$ are externally tangent at $P$, the points $O$, $P$, and $C$ are collinear.
- The distance between the centers of $\Gamma_1$ and $\Gamma_2$ is $1 + 2 = 3$.
- The distance between the centers of $\Gamma_1$ and $\Gamma_3$ is $1 + 3 = 4$.
- The distance between the centers of $\Gamma_2$ and $\Gamma_3$ is $2 + 3 = 5$.

3. Using Stewart's Theorem:
- Apply Stewart's Theorem on $\triangle ACO$ with cevian $\overline{PA}$:
$$r(r+1) + PA^2 (r+1) = r^2 + 9r$$
Simplify to find $PA^2$:
$$r(r+1) + PA^2 (r+1) = r^2 + 9r$$
$$PA^2 (r+1) = 8r$$
$$PA^2 = \frac{8r}{r+1}$$

- Apply Stewart's Theorem on $\triangle BCO$ with cevian $\overline{PB}$:
$$r(r+1) + PB^2 (r+1) = r^2 + 16r$$
Simplify to find $PB^2$:
$$r(r+1) + PB^2 (r+1) = r^2 + 16r$$
$$PB^2 (r+1) = 15r$$
$$PB^2 = \frac{15r}{r+1}$$

4. Calculate the ratio:
- The ratio $\frac{PA^2}{PB^2}$ is:
$$\frac{PA^2}{PB^2} = \frac{\frac{8r}{r+1}}{\frac{15r}{r+1}} = \frac{8r}{15r} = \frac{8}{15}$$

The final answer is $\boxed{\frac{8}{15}}$