1. Identify the given elements and their properties:
- Triangle $ABC$ with $AB = 10$, $AC = 11$, and circumradius $R = 6$.
- Points $D$ and $E$ on the circumcircle such that $\triangle ADE$ is equilateral.
- Line segments $\overline{DE}$ and $\overline{BC}$ intersect at $X$.

2. **Introduce the antipode $A'$ of $A$ with respect to the circumcircle $\odot(ABC)$:**
- The antipode $A'$ is the point on the circumcircle diametrically opposite to $A$.

3. **Calculate the distances $BA'$ and $CA'$ using the Pythagorean theorem:**
- Since $A'$ is the antipode of $A$, $AA' = 2R = 12$.
- Using the Pythagorean theorem in $\triangle ABA'$:
$$BA' = \sqrt{AB^2 + AA'^2 - 2 \cdot AB \cdot AA' \cdot \cos(\angle BAA')} = \sqrt{10^2 + 12^2 - 2 \cdot 10 \cdot 12 \cdot \cos(90^\circ)} = \sqrt{100 + 144} = \sqrt{244} = 2\sqrt{61}$$
- Similarly, in $\triangle ACA'$:
$$CA' = \sqrt{AC^2 + AA'^2 - 2 \cdot AC \cdot AA' \cdot \cos(\angle CAA')} = \sqrt{11^2 + 12^2 - 2 \cdot 11 \cdot 12 \cdot \cos(90^\circ)} = \sqrt{121 + 144} = \sqrt{265} = \sqrt{23}$$

4. **Determine the ratio $AM : MA'$ where $M = DE \cap AA'$:**
- Since $\triangle ADE$ is equilateral, $M$ divides $AA'$ in the ratio $3:1$.

5. **Introduce points $P$ and $Q$ where lines through $A$ and $A'$ perpendicular to $AA'$ intersect $BC$:**
- $P$ and $Q$ are the feet of the perpendiculars from $A$ and $A'$ to $BC$.

6. **Calculate the ratios $\frac{BP}{CP}$ and $\frac{BQ}{CQ}$:**
- Using the power of a point theorem:
$$\frac{BP}{CP} = \frac{AB^2}{AC^2} = \frac{10^2}{11^2} = \frac{100}{121}$$
$$\frac{BQ}{CQ} = \frac{A'B^2}{A'C^2} = \frac{(2\sqrt{61})^2}{(\sqrt{23})^2} = \frac{244}{23} = \frac{44}{23}$$

7. **Express $BC$ in terms of a variable $t$:**
- Let $BC = 21t$.

8. **Calculate the lengths $BP$ and $CQ$:**
- From the ratios:
$$BP = 100t \quad \text{and} \quad CQ = 23t$$

9. **Determine the length $PQ$:**
- Since $P$ and $Q$ are on $BC$:
$$PQ = 144t$$

10. **Calculate the length $PX$ using the ratio $PX : XQ = 3 : 1$:**
- Since $PX : XQ = 3 : 1$:
$$PX = \frac{3}{4}PQ = \frac{3}{4} \cdot 144t = 108t$$

11. **Determine the lengths $BX$ and $XC$:**
- Since $BX = BP + PX$:
$$BX = 100t + 108t = 208t$$
- Since $BC = 21t$:
$$XC = BC - BX = 21t - 208t = -187t$$

12. **Calculate the ratio $\frac{BX}{XC}$:**
- The ratio is:
$$\frac{BX}{XC} = \frac{208t}{-187t} = \frac{208}{-187} = -\frac{208}{187}$$

The final answer is $\boxed{\frac{8}{13}}$.