Answer: 672

Solution:

Let $O$ and $r$ be the center and radius of $\omega,$ and $O_1, O_2$ the other centers. External tangency gives $OO_1 = r + 961,$ so the power of $O$ with respect to $\omega_1$ is $OO_1^2 - 961^2 = r^2 + 2 \cdot 961r;$ similarly its power with respect to $\omega_2$ is $r^2 + 2 \cdot 625r.$ The difference is $2r(961 - 625) = 672r.$

For any point $X,$ the difference $\mathrm{pow}_{\omega_1}(X) - \mathrm{pow}_{\omega_2}(X)$ $= (XO_1^2 - XO_2^2)$ $- (961^2 - 625^2)$ is a linear function of $X$ that vanishes on the radical axis, which is line $AB;$ its rate of change perpendicular to $AB$ is $2 \cdot O_1O_2.$ So the difference equals $2 \cdot O_1O_2 \cdot \operatorname{dist}(O, AB).$ Meanwhile the chord $PQ$ of $\omega$ subtends a $120^\circ$ central angle, so $\operatorname{dist}(O, AB) = r\cos 60^\circ = \frac{r}{2}.$

Therefore $672r = 2 \cdot O_1O_2 \cdot \frac{r}{2} = O_1O_2 \cdot r,$ and the distance between the centers is $672.$