Let $h_1$ and $h_2$ be the altitudes of a triangle drawn to the sides with length $5$ and $2\sqrt 6$, respectively. If $5+h_1 \leq 2\sqrt 6 + h_2$, what is the third side of the triangle?

$ 
\textbf{(A)}\ 5
\qquad\textbf{(B)}\ 7
\qquad\textbf{(C)}\ 2\sqrt 6
\qquad\textbf{(D)}\ 3\sqrt 6
\qquad\textbf{(E)}\ 5\sqrt 3
$
1. Let $\triangle ABC$ be a triangle with sides $a = 5$, $b = 2\sqrt{6}$, and $c$ being the third side. Let $h_1$ and $h_2$ be the altitudes to the sides $a$ and $b$, respectively.
2. The given inequality is $5 + h_1 \leq 2\sqrt{6} + h_2$.
3. Recall that the area of the triangle can be expressed using the base and the corresponding altitude:
   \[
   \text{Area} = \frac{1}{2} \times a \times h_1 = \frac{1}{2} \times b \times h_2
   \]
   Therefore, we have:
   \[
   a \times h_1 = b \times h_2
   \]
   Substituting $a = 5$ and $b = 2\sqrt{6}$, we get:
   \[
   5 \times h_1 = 2\sqrt{6} \times h_2
   \]
   Solving for $h_1$ in terms of $h_2$, we get:
   \[
   h_1 = \frac{2\sqrt{6}}{5} \times h_2
   \]
4. Substitute $h_1$ back into the inequality $5 + h_1 \leq 2\sqrt{6} + h_2$:
   \[
   5 + \frac{2\sqrt{6}}{5} \times h_2 \leq 2\sqrt{6} + h_2
   \]
5. To simplify, multiply through by 5 to clear the fraction:
   \[
   25 + 2\sqrt{6} \times h_2 \leq 10\sqrt{6} + 5h_2
   \]
6. Rearrange the terms to isolate $h_2$:
   \[
   25 \leq 10\sqrt{6} + 5h_2 - 2\sqrt{6} \times h_2
   \]
   \[
   25 \leq 10\sqrt{6} + h_2(5 - 2\sqrt{6})
   \]
7. Since $h_2$ must be positive, we need to consider the geometric properties of the triangle. The inequality $5 + h_1 \leq 2\sqrt{6} + h_2$ suggests that the triangle is a right triangle. This is because the only way the inequality holds for all positive $h_1$ and $h_2$ is if the triangle is right-angled, making the altitudes proportional to the sides.
8. If $\triangle ABC$ is a right triangle, then the hypotenuse $c$ can be found using the Pythagorean theorem:
   \[
   c = \sqrt{a^2 + b^2} = \sqrt{5^2 + (2\sqrt{6})^2} = \sqrt{25 + 24} = \sqrt{49} = 7
   \]

The final answer is $\boxed{7}$