In the figure below $\triangle ABC$, $\triangle DEF$, and $\triangle GHI$ are overlapping equilateral triangles, $C$ and $F$ lie on $\overline{BD}$, $F$ and $I$ lie on $\overline{EG}$, and $C$ and $I$ lie on $\overline{AH}$. Length $AB = 2FC$, $DE = 3FC$, and $GH = 4FC$. Given that the area of $\triangle FCI$ is $3$, find the area of the hexagon $ABGHDE$.
[asy]
size(5cm);
pen dps = fontsize(10);
defaultpen(dps);
pair A,B,C,D,E,F,G,H,I;
G=origin;
H=(4,0);
I=(2,2*sqrt(3));
F=(1.5,3*sqrt(3)/2);
C=F+(1,0);
B=F-(1,0);
D=C+(2,0);
A=F+(0,sqrt(3));
E=C+(0.5,3*sqrt(3)/2);
draw(A--H--G--E--D--B--cycle);
label("$A$",A,N*.5);
label("$B$",B,S*.5);
label("$C$",C,SW*.5);
label("$D$",D,S*.5);
label("$E$",E,N*.5);
label("$F$",F,SE*.5);
label("$G$",G,S*.5);
label("$H$",H,S*.5);
label("$I$",I,N*2);
[/asy]
1. **Identify the side lengths of the triangles:**
   - Let \( FC = x \).
   - Given \( AB = 2FC \), so \( AB = 2x \).
   - Given \( DE = 3FC \), so \( DE = 3x \).
   - Given \( GH = 4FC \), so \( GH = 4x \).

2. **Calculate the areas of the equilateral triangles:**
   - The area of an equilateral triangle with side length \( s \) is given by:
     \[
     \text{Area} = \frac{\sqrt{3}}{4} s^2
     \]
   - For \( \triangle ABC \) with side length \( 2x \):
     \[
     \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} (2x)^2 = \frac{\sqrt{3}}{4} \cdot 4x^2 = \sqrt{3} x^2
     \]
   - For \( \triangle DEF \) with side length \( 3x \):
     \[
     \text{Area of } \triangle DEF = \frac{\sqrt{3}}{4} (3x)^2 = \frac{\sqrt{3}}{4} \cdot 9x^2 = \frac{9\sqrt{3}}{4} x^2
     \]
   - For \( \triangle GHI \) with side length \( 4x \):
     \[
     \text{Area of } \triangle GHI = \frac{\sqrt{3}}{4} (4x)^2 = \frac{\sqrt{3}}{4} \cdot 16x^2 = 4\sqrt{3} x^2
     \]

3. **Sum the areas of the three triangles:**
   \[
   \text{Total area of } \triangle ABC, \triangle DEF, \text{ and } \triangle GHI = \sqrt{3} x^2 + \frac{9\sqrt{3}}{4} x^2 + 4\sqrt{3} x^2
   \]
   \[
   = \sqrt{3} x^2 + 2.25\sqrt{3} x^2 + 4\sqrt{3} x^2
   \]
   \[
   = (1 + 2.25 + 4)\sqrt{3} x^2
   \]
   \[
   = 7.25\sqrt{3} x^2
   \]

4. **Subtract the overlapping areas:**
   - The area of \( \triangle FCI \) is given as 3.
   - Since \( \triangle FCI \) is counted three times in the total area calculation, we need to subtract it twice:
   \[
   \text{Adjusted total area} = 7.25\sqrt{3} x^2 - 2 \times 3
   \]
   \[
   = 7.25\sqrt{3} x^2 - 6
   \]

5. **Calculate the area of the hexagon \( ABGHDE \):**
   - The hexagon \( ABGHDE \) is formed by removing the areas of the three small triangles \( \triangle FCI \) from the combined area of the three large triangles.
   - Since the area of \( \triangle FCI \) is 3, and it is subtracted twice, the final area of the hexagon is:
   \[
   \text{Area of hexagon } ABGHDE = 7.25\sqrt{3} x^2 - 6
   \]

The final answer is \( \boxed{ 7.25\sqrt{3} x^2 - 6 } \).