A right regular hexagonal prism has bases $ABCDEF$, $A'B'C'D'E'F'$ and edges $AA'$, $BB'$, $CC'$, $DD'$, $EE'$, $FF'$, each of which is perpendicular to both hexagons. The height of the prism is $5$ and the side length of the hexagons is $6$. The plane $P$ passes through points $A$, $C'$, and $E$. The area of the portion of $P$ contained in the prism can be expressed as $m\sqrt{n}$, where $n$ is not divisible by the square of any prime. Find $m+n$.
1. **Identify the vertices and dimensions of the prism:**
   - The bases of the prism are regular hexagons $ABCDEF$ and $A'B'C'D'E'F'$.
   - The height of the prism is $5$.
   - The side length of the hexagons is $6$.

2. **Determine the coordinates of the vertices:**
   - Place the hexagon $ABCDEF$ in the $xy$-plane with center at the origin $(0,0,0)$.
   - The vertices of the hexagon can be represented as:
     \[
     A = (6, 0, 0), \quad B = (3, 3\sqrt{3}, 0), \quad C = (-3, 3\sqrt{3}, 0), \quad D = (-6, 0, 0), \quad E = (-3, -3\sqrt{3}, 0), \quad F = (3, -3\sqrt{3}, 0)
     \]
   - The corresponding vertices of the top hexagon $A'B'C'D'E'F'$ are:
     \[
     A' = (6, 0, 5), \quad B' = (3, 3\sqrt{3}, 5), \quad C' = (-3, 3\sqrt{3}, 5), \quad D' = (-6, 0, 5), \quad E' = (-3, -3\sqrt{3}, 5), \quad F' = (3, -3\sqrt{3}, 5)
     \]

3. **Find the equation of the plane $P$ passing through points $A$, $C'$, and $E$:**
   - Points: $A = (6, 0, 0)$, $C' = (-3, 3\sqrt{3}, 5)$, $E = (-3, -3\sqrt{3}, 0)$.
   - Use the determinant method to find the normal vector $\vec{n}$ of the plane:
     \[
     \vec{AC'} = (-9, 3\sqrt{3}, 5), \quad \vec{AE} = (-9, -3\sqrt{3}, 0)
     \]
     \[
     \vec{n} = \vec{AC'} \times \vec{AE} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     -9 & 3\sqrt{3} & 5 \\
     -9 & -3\sqrt{3} & 0
     \end{vmatrix} = (15\sqrt{3}, -45, 54\sqrt{3})
     \]
   - Simplify the normal vector:
     \[
     \vec{n} = 3\sqrt{3}(5, -15, 18)
     \]
   - The equation of the plane is:
     \[
     5x - 15y + 18z = d
     \]
   - Substitute point $A = (6, 0, 0)$ to find $d$:
     \[
     5(6) - 15(0) + 18(0) = 30 \implies d = 30
     \]
   - The plane equation is:
     \[
     5x - 15y + 18z = 30 \implies x - 3y + \frac{18}{5}z = 6
     \]

4. **Determine the intersection of the plane with the prism:**
   - The plane intersects the prism along a hexagonal cross-section.
   - Project the hexagon onto the $xy$-plane and find the area of the intersection.

5. **Calculate the area of the intersection:**
   - The area of the base hexagon $ABCDE$ is:
     \[
     \text{Area} = \frac{3\sqrt{3}}{2} \times s^2 = \frac{3\sqrt{3}}{2} \times 36 = 54\sqrt{3}
     \]
   - The height of the intersection is the distance from $AE$ to $C'$:
     \[
     \text{Distance} = \sqrt{(6 - (-3))^2 + (0 - 3\sqrt{3})^2 + (0 - 5)^2} = \sqrt{81 + 27 + 25} = \sqrt{133}
     \]
   - The area of the portion of $P$ contained in the prism is:
     \[
     \text{Area} = 54\sqrt{3} \times \frac{\sqrt{133}}{9} = 6\sqrt{399}
     \]

6. **Express the area in the form $m\sqrt{n}$ and find $m+n$:**
   - The area is $6\sqrt{399}$, where $m = 6$ and $n = 399$.
   - Therefore, $m+n = 6 + 399 = 405$.

The final answer is $\boxed{405}$