Lines $AD,$ $BC,$ and $XY$ are radical axes of pairs of the circles, so they meet at a point $Z$

If $M = XY \cap AB,$ then $MA^2 = MX \cdot MY = MB^2;$ show $AYBZ$ is a parallelogram and $XZ^2 = XC \cdot XD$

Line $AD$ is the radical axis of $\omega$ and $\omega_1,$ line $BC$ that of $\omega$ and $\omega_2,$ and line $XY$ that of $\omega_1$ and $\omega_2,$ so the three lines meet at the radical center $Z.$ (They cannot be parallel: that would force a symmetric configuration with $XC = XD.$) Let $M = XY \cap AB.$ The power of $M$ with respect to each circle gives $MA^2 = MX \cdot MY = MB^2,$ so $M$ is the midpoint of $\overline{AB},$ with $X$ between $M$ and $Y.$

Since $ADYX$ is cyclic, $\angle XAZ = \angle XYD,$ and since $BCYX$ is cyclic, $\angle XBZ = \angle XYC;$ as $C,$ $Y,$ $D$ are collinear these add to $180^\circ,$ so $ZAXB$ is cyclic. The tangent-chord angle at $B$ gives $\angle XYB = \angle ABX = \angle AZX,$ so $BY \parallel ZA,$ and symmetrically $AY \parallel ZB.$ Hence $AYBZ$ is a parallelogram, and since $M$ is the midpoint of diagonal $\overline{AB},$ it is also the midpoint of $\overline{ZY}:$ therefore $XZ = XM + MZ$ $= MX + MY.$ Moreover $\angle XCZ = \angle XYB = \angle XZD$ and (by the tangent-chord angle at $A$) $\angle XZC = \angle XAB = \angle XYA$ $= \angle XDZ,$ so triangles $XZC$ and $XDZ$ are similar, giving $XZ^2 = XC \cdot XD.$

Putting it together, $$\begin{aligned} AB^2 &= 4MA^2 = 4\,MX \cdot MY \\ &= (MX + MY)^2 \\ &\quad {}- (MY - MX)^2 \\ &= XZ^2 - XY^2 \\ &= XC \cdot XD - XY^2, \end{aligned}$$ which equals $67 \cdot 37 - 47^2 = 2479 - 2209$ $= 270.$