Consider an isosceles triangle $ABC$ with side lengths $AB = AC = 10\sqrt{2}$ and $BC =10\sqrt{3}$.  Construct semicircles $P$, $Q$, and $R$ with diameters $AB$, $AC$, $BC$ respectively, such that the plane of each semicircle is perpendicular to the plane of $ABC$, and all semicircles are on the same side of plane $ABC$ as shown.  There exists a plane above triangle $ABC$ that is tangent to all three semicircles $P$, $Q$, $R$ at the points $D$, $E$, and $F$ respectively, as shown in the diagram.  Calculate, with proof, the area of triangle $DEF$.
[asy]
size(200);
import three; defaultpen(linewidth(0.7)+fontsize(10)); currentprojection = orthographic(0,4,2.5); 
// 1.15 x-scale distortion factor 
triple A = (0,0,0), B = (75^.5/1.15,-125^.5,0), C = (-75^.5/1.15,-125^.5,0), D = (A+B)/2 + (0,0,abs((B-A)/2)), E = (A+C)/2 + (0,0,abs((C-A)/2)), F = (C+B)/2 + (0,0,abs((B-C)/2)); 
draw(D--E--F--cycle); draw(B--A--C); 
// approximate guess for r
real r = 1.38; draw(B--(r*B+C)/(1+r)^^(B+r*C)/(1+r)--C,linetype("4 4")); draw((B+r*C)/(1+r)--(r*B+C)/(1+r));
// lazy so I'll draw six arcs
draw(arc((A+B)/2,A,D)); draw(arc((A+B)/2,D,B)); draw(arc((A+C)/2,E,A)); draw(arc((A+C)/2,E,C)); draw(arc((C+B)/2,F,B)); draw(arc((C+B)/2,F,C)); 
label("$A$",A,S); label("$B$",B,W); label("$C$",C,plain.E);
label("$D$",D,SW); label("$E$",E,SE); label("$F$",F,N);[/asy]
1. **Identify the Geometry and Setup:**
   - Given an isosceles triangle \(ABC\) with \(AB = AC = 10\sqrt{2}\) and \(BC = 10\sqrt{3}\).
   - Semicircles \(P\), \(Q\), and \(R\) are constructed with diameters \(AB\), \(AC\), and \(BC\) respectively, perpendicular to the plane of \(ABC\).
   - A plane above \(ABC\) is tangent to all three semicircles at points \(D\), \(E\), and \(F\).

2. **Determine the Radii of the Semicircles:**
   - Radius of semicircle \(P\) (diameter \(AB\)): \(r_P = \frac{AB}{2} = 5\sqrt{2}\).
   - Radius of semicircle \(Q\) (diameter \(AC\)): \(r_Q = \frac{AC}{2} = 5\sqrt{2}\).
   - Radius of semicircle \(R\) (diameter \(BC\)): \(r_R = \frac{BC}{2} = 5\sqrt{3}\).

3. **Calculate the Heights of the Tangent Points:**
   - The tangent points \(D\), \(E\), and \(F\) are vertically above the midpoints of \(AB\), \(AC\), and \(BC\) respectively.
   - Height of \(D\) above \(AB\): \(h_D = 5\sqrt{2}\).
   - Height of \(E\) above \(AC\): \(h_E = 5\sqrt{2}\).
   - Height of \(F\) above \(BC\): \(h_F = 5\sqrt{3}\).

4. **Determine the Coordinates of the Tangent Points:**
   - Midpoint of \(AB\): \(M_{AB} = \left(\frac{A+B}{2}\right)\).
   - Midpoint of \(AC\): \(M_{AC} = \left(\frac{A+C}{2}\right)\).
   - Midpoint of \(BC\): \(M_{BC} = \left(\frac{B+C}{2}\right)\).
   - Coordinates of \(D\): \(D = M_{AB} + (0, 0, 5\sqrt{2})\).
   - Coordinates of \(E\): \(E = M_{AC} + (0, 0, 5\sqrt{2})\).
   - Coordinates of \(F\): \(F = M_{BC} + (0, 0, 5\sqrt{3})\).

5. **Calculate the Lengths of \(DE\) and \(DF\):**
   - Since \(D\) and \(E\) are at the same height, \(DE\) is the horizontal distance between \(M_{AB}\) and \(M_{AC}\).
   - Similarly, \(DF\) is the horizontal distance between \(M_{AB}\) and \(M_{BC}\).

6. **Use Similar Triangles to Find \(DE\) and \(DF\):**
   - By symmetry and properties of similar triangles:
     \[
     \frac{DE}{GH} = \frac{IJ}{IF}
     \]
   - Given \(IJ = \frac{8}{3}\sqrt{3}\) and \(JF = 4\sqrt{3}\), we can find \(DE\).

7. **Calculate the Area of \(\triangle DEF\):**
   - Using the formula for the area of a triangle:
     \[
     \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
     \]
   - Here, base \(DE = 4\sqrt{3}\) and height \(JF = 4\sqrt{3}\):
     \[
     \text{Area} = \frac{1}{2} \times 4\sqrt{3} \times 4\sqrt{3} = 24
     \]

The final answer is \(\boxed{24}\).