Solution:

Write $s = \sqrt{111}$ and $r = \sqrt{11}.$ Since each triangle $AD_iE_k$ is congruent to $\triangle ABC,$ we have $AD_i = AE_k = s,$ so $D_1$ and $D_2$ are the two intersections of the circle of radius $s$ about $A$ with the circle of radius $r$ about $B;$ they are mirror images across line $AB,$ so $\angle BAD_1 = \angle BAD_2 = \theta$ with $D_1$ and $D_2$ on opposite sides of $AB.$ Each $E_k$ is the image of its $D_i$ rotated $\pm 60^\circ$ about $A.$ Measuring signed angles from ray $AB,$ with $C$ at $+60^\circ,$ the rays $AD_i$ sit at $\pm\theta$ and the rays $AE_k$ at $\pm\theta \pm 60^\circ,$ so the four angles $\angle CAE_k$ are $\theta,$ $\theta,$ $120^\circ - \theta,$ and $120^\circ + \theta.$

Since $AC = AE_k = s,$ the law of cosines gives $(CE_k)^2 = 2s^2(1 - \cos\angle CAE_k).$ Using $\cos(120^\circ - \theta) + \cos(120^\circ + \theta) = 2\cos 120^\circ \cos\theta = -\cos\theta,$ the four angles' cosines sum to $2\cos\theta - \cos\theta = \cos\theta,$ so $$\sum_{k=1}^{4}(CE_k)^2 = 2s^2(4 - \cos\theta).$$

Applying the law of cosines in triangle $ABD_1$ (with $AB = AD_1 = s$) gives $r^2 = 2s^2(1 - \cos\theta),$ so $2s^2\cos\theta = 2s^2 - r^2.$ Therefore the sum equals $8s^2 - (2s^2 - r^2) = 6s^2 + r^2 = 6 \cdot 111 + 11 = 677.$