Define a common chord between two intersecting circles to be the line segment connecting their two intersection points. Let $\omega_1,\omega_2,\omega_3$ be three circles of radii $3, 5,$ and $7$, respectively. Suppose they are arranged in such a way that the common chord of $\omega_1$ and $\omega_2$ is a diameter of $\omega_1$, the common chord of $\omega_1$ and $\omega_3$ is a diameter of $\omega_1$, and the common chord of $\omega_2$ and $\omega_3$ is a diameter of $\omega_2$. Compute the square of the area of the triangle formed by the centers of the three circles.

1. **Identify the centers and radii of the circles:**
   - Let the centers of the circles $\omega_1, \omega_2, \omega_3$ be $O_1, O_2, O_3$ respectively.
   - The radii of the circles are $r_1 = 3$, $r_2 = 5$, and $r_3 = 7$.

2. **Determine the distances between the centers:**
   - The common chord of $\omega_1$ and $\omega_2$ is a diameter of $\omega_1$. Therefore, the distance between $O_1$ and $O_2$ is $2 \times r_1 = 2 \times 3 = 6$.
   - The common chord of $\omega_1$ and $\omega_3$ is a diameter of $\omega_1$. Therefore, the distance between $O_1$ and $O_3$ is $2 \times r_1 = 2 \times 3 = 6$.
   - The common chord of $\omega_2$ and $\omega_3$ is a diameter of $\omega_2$. Therefore, the distance between $O_2$ and $O_3$ is $2 \times r_2 = 2 \times 5 = 10$.

3. **Verify the triangle formed by the centers:**
   - The lengths of the sides of the triangle formed by $O_1, O_2, O_3$ are $O_1O_2 = 6$, $O_1O_3 = 6$, and $O_2O_3 = 10$.
   - Check if the triangle is a right triangle using the Pythagorean theorem:
     \[
     O_1O_2^2 + O_1O_3^2 = 6^2 + 6^2 = 36 + 36 = 72
     \]
     \[
     O_2O_3^2 = 10^2 = 100
     \]
     Since $72 \neq 100$, the triangle is not a right triangle. Therefore, the initial solution's assumption of a right triangle is incorrect.

4. **Calculate the area of the triangle using Heron's formula:**
   - Let $a = 6$, $b = 6$, and $c = 10$.
   - Compute the semi-perimeter $s$:
     \[
     s = \frac{a + b + c}{2} = \frac{6 + 6 + 10}{2} = 11
     \]
   - Use Heron's formula to find the area $A$:
     \[
     A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{11(11-6)(11-6)(11-10)} = \sqrt{11 \cdot 5 \cdot 5 \cdot 1} = \sqrt{275}
     \]

5. **Compute the square of the area:**
   \[
   (\text{Area})^2 = (\sqrt{275})^2 = 275
   \]

The final answer is $\boxed{275}$.