In the star shaped figure below, if all side lengths are equal to $3$ and the three largest angles of the figure are $210$ degrees, its area can be expressed as $\frac{a \sqrt{b}}{c}$ , where $a, b$, and $c$ are positive integers such that $a$ and $c$ are relatively prime and that $b$ is square-free. Compute $a + b + c$.
[img]https://cdn.artofproblemsolving.com/attachments/a/f/d16a78317b0298d6894c6bd62fbcd1a5894306.png[/img]
1. **Connect the three "pointy" vertices to form an equilateral triangle:**
   - Let the side length of the equilateral triangle be \( s \).
   - Use the Law of Cosines to find \( s \).

2. **Apply the Law of Cosines:**
   \[
   s^2 = 3^2 + 3^2 - 2 \cdot 3 \cdot 3 \cdot \cos(150^\circ)
   \]
   - Note that \(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\).
   \[
   s^2 = 9 + 9 - 18 \cdot \left(-\frac{\sqrt{3}}{2}\right)
   \]
   \[
   s^2 = 18 + 9\sqrt{3}
   \]

3. **Calculate the area of the equilateral triangle:**
   - The area \( A \) of an equilateral triangle with side length \( s \) is given by:
   \[
   A = \frac{s^2 \sqrt{3}}{4}
   \]
   - Substitute \( s^2 = 18 + 9\sqrt{3} \):
   \[
   A = \frac{18 + 9\sqrt{3}}{4} \cdot \sqrt{3}
   \]
   \[
   A = \frac{18\sqrt{3} + 27}{4}
   \]

4. **Calculate the area of the three smaller triangles:**
   - Each smaller triangle has a base of 3 and a height determined by the sine of 150 degrees.
   - The area of one smaller triangle is:
   \[
   \text{Area} = \frac{1}{2} \cdot 3 \cdot 3 \cdot \sin(150^\circ)
   \]
   - Note that \(\sin(150^\circ) = \frac{1}{2}\):
   \[
   \text{Area} = \frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{1}{2} = \frac{9}{2}
   \]
   - The total area of the three smaller triangles is:
   \[
   3 \cdot \frac{9}{2} = \frac{27}{2}
   \]

5. **Calculate the area of the star-shaped figure:**
   - Subtract the area of the three smaller triangles from the area of the equilateral triangle:
   \[
   \text{Area} = \frac{18\sqrt{3} + 27}{4} - \frac{27}{2}
   \]
   - Convert \(\frac{27}{2}\) to a common denominator:
   \[
   \frac{27}{2} = \frac{54}{4}
   \]
   - Subtract the fractions:
   \[
   \text{Area} = \frac{18\sqrt{3} + 27 - 54}{4} = \frac{18\sqrt{3} - 27}{4}
   \]

6. **Express the area in the form \(\frac{a \sqrt{b}}{c}\):**
   - The area is \(\frac{9\sqrt{3}}{2}\), where \(a = 9\), \(b = 3\), and \(c = 2\).
   - Verify that \(a\) and \(c\) are relatively prime and \(b\) is square-free.

The final answer is \(9 + 3 + 2 = \boxed{14}\).