Equilateral triangle $ T$ is inscribed in circle $ A$, which has radius $ 10$. Circle $ B$ with radius $ 3$ is internally tangent to circle $ A$ at one vertex of $ T$. Circles $ C$ and $ D$, both with radius $ 2$, are internally tangent to circle $ A$ at the other two vertices of $ T$. Circles $ B$, $ C$, and $ D$ are all externally tangent to circle $ E$, which has radius $ \frac {m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$.
[asy]unitsize(2.2mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=4;

pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90);
pair Ep=(0,4-27/5);
pair[] dotted={A,B,C,D,Ep};

draw(Circle(A,10));
draw(Circle(B,3));
draw(Circle(C,2));
draw(Circle(D,2));
draw(Circle(Ep,27/5));

dot(dotted);
label("$E$",Ep,E);
label("$A$",A,W);
label("$B$",B,W);
label("$C$",C,W);
label("$D$",D,E);[/asy]
1. Let the radius of circle \( E \) be \( x \). We need to find \( x \) such that circles \( B \), \( C \), and \( D \) are all externally tangent to circle \( E \).

2. Consider the geometry of the problem. Circle \( A \) has a radius of 10, and the equilateral triangle \( T \) is inscribed in circle \( A \). The side length of \( T \) can be found using the formula for the side length of an equilateral triangle inscribed in a circle:
   \[
   s = R \sqrt{3}
   \]
   where \( R \) is the radius of the circumscribed circle. Thus,
   \[
   s = 10 \sqrt{3}
   \]

3. Circle \( B \) with radius 3 is internally tangent to circle \( A \) at one vertex of \( T \). Circles \( C \) and \( D \) with radius 2 are internally tangent to circle \( A \) at the other two vertices of \( T \).

4. Let’s denote the vertices of the equilateral triangle \( T \) as \( P \), \( Q \), and \( R \). Circle \( B \) is tangent to circle \( A \) at vertex \( P \), circle \( C \) is tangent to circle \( A \) at vertex \( Q \), and circle \( D \) is tangent to circle \( A \) at vertex \( R \).

5. We need to find the radius \( x \) of circle \( E \) such that it is externally tangent to circles \( B \), \( C \), and \( D \).

6. Consider the distances between the centers of these circles. The distance between the center of circle \( A \) and the center of circle \( B \) is:
   \[
   10 - 3 = 7
   \]
   The distance between the center of circle \( A \) and the center of circle \( C \) or \( D \) is:
   \[
   10 - 2 = 8
   \]

7. Now, consider the distances from the center of circle \( E \) to the centers of circles \( B \), \( C \), and \( D \). Let’s denote the center of circle \( E \) as \( O_E \), the center of circle \( B \) as \( O_B \), the center of circle \( C \) as \( O_C \), and the center of circle \( D \) as \( O_D \).

8. The distance \( O_E O_B \) is:
   \[
   x + 3
   \]
   The distance \( O_E O_C \) and \( O_E O_D \) is:
   \[
   x + 2
   \]

9. Using the Law of Cosines in triangle \( O_A O_C O_E \) (where \( O_A \) is the center of circle \( A \)), we have:
   \[
   (x + 2)^2 = (x - 4)^2 + 64 - 2 \cdot 8 \cdot (x - 4) \cdot \cos(60^\circ)
   \]
   Since \( \cos(60^\circ) = \frac{1}{2} \), the equation simplifies to:
   \[
   (x + 2)^2 = (x - 4)^2 + 64 - 8(x - 4)
   \]

10. Expanding and simplifying the equation:
    \[
    x^2 + 4x + 4 = x^2 - 8x + 16 + 64 - 8x + 32
    \]
    \[
    x^2 + 4x + 4 = x^2 - 16x + 112
    \]
    \[
    4x + 4 = -16x + 112
    \]
    \[
    20x = 108
    \]
    \[
    x = \frac{108}{20} = \frac{27}{5}
    \]

11. Therefore, the radius of circle \( E \) is \( \frac{27}{5} \). The values of \( m \) and \( n \) are 27 and 5, respectively.

12. The sum \( m + n \) is:
    \[
    27 + 5 = 32
    \]

The final answer is \( \boxed{32} \)