1. Identify Key Elements and Relationships:
- Given: $\triangle ABC$ with $BC = 4\sqrt{6}$.
- $\omega$ is the circumcircle of $\triangle ABC$.
- $AD$ is the diameter of $\omega$.
- $N$ is the midpoint of arc $BC$ that contains $A$.
- $H$ is the orthocenter of $\triangle ABC$.
- $HN = HD = 6$.

2. Establishing the Geometry:
- Let $O$ be the circumcenter of $\triangle ABC$.
- Since $AD$ is the diameter, $D$ is the antipode of $A$.
- $N$ is the midpoint of the arc $BC$ that contains $A$, so $N$ lies on the perpendicular bisector of $BC$ and $ON \perp BC$.
- $H$ is the orthocenter, so $H$ lies on the altitudes of $\triangle ABC$.

3. Using the Isosceles Triangle Property:
- $\triangle HND$ is isosceles with $HN = HD = 6$.
- Let $M$ be the midpoint of $ND$. Since $\triangle HND$ is isosceles, $HM \perp ND$.
- $O$ is the circumcenter, so $OM \perp ND$.
- Therefore, $H, O, M$ are collinear.

4. **Proving $O$ is the Centroid of $\triangle HND$:**
- $P$ is the midpoint of $BC$.
- $N$ is the midpoint of arc $BC$, so $N, O, P$ are collinear.
- Since $H, O, M$ are collinear and $O$ lies on the line $NP$, $O$ is the centroid of $\triangle HND$.

5. **Finding the Circumradius $R$:**
- Since $N$ is the midpoint of arc $BC$, $NO = 2OP$.
- $OP = \frac{R}{2}$.
- $OB = R$, so $\triangle OBP$ is a 30-60-90 triangle.
- Therefore, $\angle A = 60^\circ$.
- Using $BC = 4\sqrt{6}$, $R = \frac{BC}{\sqrt{3}} = 4\sqrt{2}$.

6. **Proving $AHON$ is a Parallelogram:**
- $DO \cap HN = Q$ is the midpoint of $HN$.
- $OD = R$, $OQ = \frac{R}{2}$, and $AO = R$.
- $Q$ bisects $AO$, so $AHON$ is a parallelogram.

7. **Calculating $OH$:**
- Let $OM = x$.
- $\triangle DMO \sim \triangle DNA$, so $AN = 2x$.
- $OH = 2x$.
- Using $HN^2 - HM^2 = NO^2 - OM^2$, we find $x = \frac{\sqrt{2}}{2}$.
- Therefore, $OH = \sqrt{2}$.

8. Using Power of a Point:
- $AHON$ is a parallelogram, so $AH = ON = R = 4\sqrt{2}$.
- Let $AH$ intersect $BC$ at $X$ and $\omega$ again at $Y \neq A$.
- $Y$ is the reflection of $H$ about $X$, so $HX = XY$.
- By Power of a Point, $R^2 - OH^2 = AH \cdot HY = AH \cdot 2HX$.
- Solving, $HX = \frac{15}{4\sqrt{2}}$.
- Thus, $AX = AH + HX = \frac{47}{4\sqrt{2}}$.

9. **Finding the Area of $\triangle ABC$:**
- $[ABC] = \frac{AD \cdot BC}{2} = \frac{47\sqrt{3}}{2}$.
- Therefore, $a = 47$, $b = 3$, $c = 2$.
- $a + b + c = 47 + 3 + 2 = 52$.

The final answer is $\boxed{52}$.