In triangle $ABC$, let $M$ be the midpoint of $BC$, $H$ be the orthocenter, and $O$ be the circumcenter. Let $N$ be the reflection of $M$ over $H$. Suppose that $OA = ON = 11$ and $OH = 7.$ Compute $BC^2$.
1. **Identify Key Points and Properties:**
   - Let \( M \) be the midpoint of \( BC \).
   - Let \( H \) be the orthocenter of \( \triangle ABC \).
   - Let \( O \) be the circumcenter of \( \triangle ABC \).
   - Let \( N \) be the reflection of \( M \) over \( H \).
   - Given: \( OA = ON = 11 \) and \( OH = 7 \).

2. **Reflection Property:**
   - Since \( N \) is the reflection of \( M \) over \( H \), we have \( NH = HM \).

3. **Midpoint Property:**
   - Let \( H' \) be the reflection of \( H \) over \( M \). It is known that \( H' \) lies on the circumcircle of \( \triangle ABC \).

4. **Distance Calculation:**
   - Since \( N \) is the reflection of \( M \) over \( H \), \( N \) and \( H' \) are the same point. Therefore, \( N \) lies on the circumcircle of \( \triangle ABC \).

5. **Distance from Circumcenter:**
   - Given \( OA = ON = 11 \), and since \( N \) lies on the circumcircle, \( ON = R \), where \( R \) is the circumradius of \( \triangle ABC \).

6. **Orthocenter and Circumcenter Relationship:**
   - Given \( OH = 7 \), we need to find \( OM \). Since \( M \) is the midpoint of \( BC \), \( OM \) is the distance from the circumcenter to the midpoint of one side of the triangle.

7. **Right Triangle Property:**
   - In \( \triangle OMB \), \( OM \) is perpendicular to \( BC \) because \( O \) is the circumcenter and \( M \) is the midpoint of \( BC \). Therefore, \( \triangle OMB \) is a right triangle with \( OM \) as one leg and \( OB \) as the hypotenuse.

8. **Pythagorean Theorem:**
   - Using the Pythagorean theorem in \( \triangle OMB \):
     \[
     OB^2 = OM^2 + BM^2
     \]
     Given \( OB = R = 11 \) and \( OM = OH = 7 \):
     \[
     11^2 = 7^2 + BM^2
     \]
     \[
     121 = 49 + BM^2
     \]
     \[
     BM^2 = 121 - 49
     \]
     \[
     BM^2 = 72
     \]
     \[
     BM = \sqrt{72} = 6\sqrt{2}
     \]

9. **Length of \( BC \):**
   - Since \( M \) is the midpoint of \( BC \), \( BC = 2BM \):
     \[
     BC = 2 \times 6\sqrt{2} = 12\sqrt{2}
     \]

10. **Square of \( BC \):**
    \[
    BC^2 = (12\sqrt{2})^2 = 144 \times 2 = 288
    \]

Conclusion:
\[
BC^2 = \boxed{288}
\]