Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$,
with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$.
Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect.  Let $D_1$ be the point on the opposite side of $\overline{AB}$ from $C$, and $D_2$ the point on the same side of $\overline{AB}$ as $C$.
Let $\theta$ be the measure of angle $BAD_1$ (also the measure of angle $BAD_2$); by the Law of Cosines,
\begin{align*}\sqrt{11}^2 &= \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot \cos\theta\\ 11 &= 222(1 - \cos\theta)\end{align*}
There are two equilateral triangles with $\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$, and $E_2$ be the third vertex that is nearest to $C$.
Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta$; by the Law of Cosines,
\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) (AC)\cos(120 + \theta)\\ &= 111 + 111 - 222\cos(120 + \theta)\end{align*}
Angle $E_2AC = \theta$; by the Law of Cosines,
\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) (AC)\cos\theta\\ &= 111 + 111 - 222\,\cos\theta\end{align*}
There are two equilateral triangles with $\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$, and $E_4$ be the third vertex that is nearest to $C$.
Angle $E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta$; by the Law of Cosines,
\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) (AC)\cos(120 - \theta)\\ &= 111 + 111 - 222\cos(120 - \theta)\end{align*}
Angle $E_4AC = \theta$; by the Law of Cosines,
\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) (AC)\cos\theta \\ &= 111 + 111 - 222\cos\theta\end{align*}
The solution is:
\begin{align*} \sum_{k=1}^4(CE_k)^2 &= (E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2\\ &= 222(1 - \cos(120 + \theta)) + 222(1 - \cos(120 - \theta)) + 222(1 - \cos\theta) + 222(1 - \cos\theta)\\ &= 222((1 - (\cos120\cos\theta - \sin120\sin\theta)) + (1 - (\cos120\cos\theta + \sin120\sin\theta)) + 2(1 -\cos\theta))\\ &= 222(1 - \cos120\cos\theta + \sin120\sin\theta + 1 - \cos120\cos\theta - \sin120\sin\theta + 2 - 2\cos\theta)\\ &= 222(1 + \frac{1}{2}\cos\theta + 1 + \frac{1}{2}\cos\theta + 2 - 2\cos\theta)\\ &= 222(4 - \cos\theta)\\ &= 666 + 222(1 - \cos\theta) \end{align*}
Substituting $11$ for $222(1 - \cos\theta)$ gives the solution $666 + 11 = \framebox{677}.$