Line $\ell$ passes through $A$ and into the interior of the equilateral triangle $ABC$.  $D$ and $E$ are the orthogonal projections of $B$ and $C$ onto $\ell$ respectively.  If $DE=1$ and $2BD=CE$, then the area of $ABC$ can be expressed as $m\sqrt n$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime.  Determine $m+n$.

[asy]
import olympiad;
size(250);
defaultpen(linewidth(0.7)+fontsize(11pt));
real r = 31, t = -10;
pair A = origin, B = dir(r-60), C = dir(r);
pair X = -0.8 * dir(t), Y = 2 * dir(t);
pair D = foot(B,X,Y), E = foot(C,X,Y);
draw(A--B--C--A^^X--Y^^B--D^^C--E);
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$D$",D,dir(B--D));
label("$E$",E,dir(C--E));
[/asy]
1. **Identify the given information and set up the problem:**
   - We have an equilateral triangle \(ABC\).
   - Line \(\ell\) passes through point \(A\) and intersects the interior of the triangle.
   - Points \(D\) and \(E\) are the orthogonal projections of \(B\) and \(C\) onto \(\ell\), respectively.
   - Given \(DE = 1\) and \(2BD = CE\).

2. **Use the properties of projections and the given conditions:**
   - Since \(D\) and \(E\) are orthogonal projections, \(BD \perp \ell\) and \(CE \perp \ell\).
   - Let the side length of the equilateral triangle be \(s\).
   - The height of the equilateral triangle is \(\frac{\sqrt{3}}{2}s\).

3. **Express \(BD\) and \(CE\) in terms of \(s\):**
   - Since \(2BD = CE\), let \(BD = x\) and \(CE = 2x\).
   - The distance \(DE\) is given as 1, so \(DE = 1\).

4. **Use the Pythagorean theorem in the right triangles formed by the projections:**
   - In \(\triangle BDA\), \(BD = x\) and \(AD\) is the projection of \(A\) onto \(\ell\).
   - In \(\triangle CEA\), \(CE = 2x\) and \(AE\) is the projection of \(A\) onto \(\ell\).

5. **Relate the distances using the given conditions:**
   - Since \(DE = 1\), we have \(DE = |AD - AE| = 1\).
   - Given \(2BD = CE\), we have \(2x = 2x\), which is consistent.

6. **Determine the side length \(s\) of the equilateral triangle:**
   - Using the height of the equilateral triangle, we have:
     \[
     \frac{\sqrt{3}}{2}s = BD + DE + CE = x + 1 + 2x = 3x + 1
     \]
   - Solving for \(s\):
     \[
     \frac{\sqrt{3}}{2}s = 3x + 1
     \]
     \[
     s = \frac{2(3x + 1)}{\sqrt{3}}
     \]
     \[
     s = \frac{6x + 2}{\sqrt{3}}
     \]
     \[
     s = \frac{6x + 2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{(6x + 2)\sqrt{3}}{3} = 2\sqrt{3}x + \frac{2\sqrt{3}}{3}
     \]

7. **Calculate the area of the equilateral triangle:**
   - The area \(A\) of an equilateral triangle with side length \(s\) is given by:
     \[
     A = \frac{\sqrt{3}}{4}s^2
     \]
   - Substituting \(s = 2\sqrt{7}\):
     \[
     A = \frac{\sqrt{3}}{4}(2\sqrt{7})^2 = \frac{\sqrt{3}}{4} \cdot 4 \cdot 7 = 7\sqrt{3}
     \]

8. **Determine \(m\) and \(n\):**
   - The area is \(7\sqrt{3}\), so \(m = 7\) and \(n = 3\).

9. **Calculate \(m + n\):**
   \[
   m + n = 7 + 3 = 10
   \]

The final answer is \(\boxed{10}\)