 Triangle $ABC$ is so that $AB = 15$, $BC = 22$, and $AC = 20$. Let $D, E, F$ lie on $BC$, $AC$, and $AB$, respectively, so $AD$, $BE$, $CF$ all contain a point $K$. Let $L$ be the second intersection of the circumcircles of $BFK$ and $CEK$. Suppose that $\frac{AK}{KD} = \frac{11}{7}$ , and $BD = 6$. If $KL^2 =\frac{a}{b}$, where $a, b$ are relatively prime integers, find $a + b$.

1. **Using Menelaus' Theorem:**
   Menelaus' theorem states that for a triangle \( \triangle ABC \) with a transversal line intersecting \( BC, CA, \) and \( AB \) at points \( D, E, \) and \( F \) respectively, the following relation holds:
   \[
   \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1
   \]
   Given \( \frac{AK}{KD} = \frac{11}{7} \), we can write:
   \[
   \frac{AK}{KD} = \frac{11}{7} \implies \frac{KD}{AK} = \frac{7}{11}
   \]

2. **Finding \( \frac{AE}{EC} \):**
   Using Menelaus' theorem on \( \triangle ABC \) with transversal \( ADE \):
   \[
   \frac{CB}{BD} \cdot \frac{DK}{KA} \cdot \frac{AE}{EC} = 1
   \]
   Given \( BD = 6 \) and \( BC = 22 \):
   \[
   \frac{22}{6} \cdot \frac{7}{11} \cdot \frac{AE}{EC} = 1 \implies \frac{11}{3} \cdot \frac{7}{11} \cdot \frac{AE}{EC} = 1 \implies \frac{7}{3} \cdot \frac{AE}{EC} = 1 \implies \frac{AE}{EC} = \frac{3}{7}
   \]

3. **Finding \( \frac{AF}{FB} \):**
   Using Menelaus' theorem on \( \triangle ABC \) with transversal \( ADF \):
   \[
   \frac{BC}{CD} \cdot \frac{DK}{KA} \cdot \frac{AF}{FB} = 1
   \]
   Given \( CD = 16 \) (since \( BD = 6 \) and \( BC = 22 \)):
   \[
   \frac{22}{16} \cdot \frac{7}{11} \cdot \frac{AF}{FB} = 1 \implies \frac{11}{8} \cdot \frac{7}{11} \cdot \frac{AF}{FB} = 1 \implies \frac{7}{8} \cdot \frac{AF}{FB} = 1 \implies \frac{AF}{FB} = \frac{8}{7}
   \]

4. **Using the Radical Axis Theorem:**
   The radical axis theorem states that if three circles intersect pairwise, then the radical axes of each pair of circles are concurrent. Here, we use the fact that \( A \) lies on the radical axis of the circumcircles of \( \triangle BFK \) and \( \triangle CEK \):
   \[
   AF \cdot AB = AE \cdot AC = 120
   \]
   This implies:
   \[
   AK \cdot AL = 120
   \]

5. **Applying Stewart's Theorem:**
   Stewart's theorem for triangle \( \triangle ABC \) with cevian \( AD \) states:
   \[
   AB^2 \cdot CD + AC^2 \cdot BD = (BD + CD)(AD^2 + BD \cdot CD)
   \]
   Plugging in the given values:
   \[
   15^2 \cdot 16 + 20^2 \cdot 6 = (6 + 16)(AD^2 + 6 \cdot 16)
   \]
   Simplifying:
   \[
   3600 + 2400 = 22(AD^2 + 96) \implies 6000 = 22AD^2 + 2112 \implies 3888 = 22AD^2 \implies AD^2 = \frac{3888}{22} = 176.727 \implies AD = \sqrt{176.727} = \frac{18\sqrt{66}}{11}
   \]

6. **Finding \( AK \):**
   Given \( \frac{AK}{KD} = \frac{11}{7} \):
   \[
   AK = \frac{11}{18} \cdot AD = \frac{11}{18} \cdot \frac{18\sqrt{66}}{11} = \sqrt{66}
   \]

7. **Finding \( KL \):**
   Let \( KL = x \):
   \[
   \sqrt{66} \cdot (\sqrt{66} + x) = 120 \implies 66 + \sqrt{66}x = 120 \implies \sqrt{66}x = 54 \implies x = \frac{54}{\sqrt{66}} = \frac{54\sqrt{66}}{66} = \frac{9\sqrt{66}}{11}
   \]
   Squaring \( x \):
   \[
   x^2 = \left(\frac{9\sqrt{66}}{11}\right)^2 = \frac{81 \cdot 66}{121} = \frac{5346}{121} = \frac{486}{11}
   \]

The final answer is \( \boxed{497} \)