A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths $3$, $4$, and $5$. What is the area of the triangle?
$\textbf{(A)}\ 6 \qquad
\textbf{(B)}\ \frac{18}{\pi^2} \qquad
\textbf{(C)}\ \frac{9}{\pi^2}\left(\sqrt{3}-1\right) \qquad
\textbf{(D)}\ \frac{9}{\pi^2}\left(\sqrt{3}+1\right) \qquad
\textbf{(E)}\ \frac{9}{\pi^2}\left(\sqrt{3}+3\right)$
1. **Determine the radius of the circle:**
   The perimeter of the circle is the sum of the lengths of the arcs, which is \(3 + 4 + 5 = 12\). The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius. Therefore, we have:
   \[
   2 \pi r = 12 \implies r = \frac{6}{\pi}
   \]

2. **Determine the central angles corresponding to each arc:**
   The length of an arc is proportional to the central angle it subtends. The total circumference corresponds to \(360^\circ\). Therefore, the angles corresponding to the arcs of lengths 3, 4, and 5 are:
   \[
   \text{Angle for arc 3: } \frac{3}{12} \times 360^\circ = 90^\circ
   \]
   \[
   \text{Angle for arc 4: } \frac{4}{12} \times 360^\circ = 120^\circ
   \]
   \[
   \text{Angle for arc 5: } \frac{5}{12} \times 360^\circ = 150^\circ
   \]

3. **Use the Law of Cosines to find the sides of the triangle:**
   For each central angle, we can use the Law of Cosines to find the corresponding side of the triangle. The Law of Cosines states:
   \[
   c^2 = a^2 + b^2 - 2ab \cos(\theta)
   \]
   where \(a\) and \(b\) are the radii of the circle (both equal to \(r\)), and \(\theta\) is the central angle.

   - For the angle \(90^\circ\):
     \[
     a^2 = r^2 + r^2 - 2r^2 \cos(90^\circ) = 2r^2 \implies a = \sqrt{2} r
     \]

   - For the angle \(120^\circ\):
     \[
     b^2 = r^2 + r^2 - 2r^2 \cos(120^\circ) = 2r^2 (1 - \cos(120^\circ)) = 2r^2 (1 + \frac{1}{2}) = 3r^2 \implies b = \sqrt{3} r
     \]

   - For the angle \(150^\circ\):
     \[
     c^2 = r^2 + r^2 - 2r^2 \cos(150^\circ) = 2r^2 (1 - \cos(150^\circ)) = 2r^2 (1 + \frac{\sqrt{3}}{2}) = (2 + \sqrt{3})r^2 \implies c = \sqrt{(2 + \sqrt{3})} r
     \]

4. **Calculate the area of the triangle using the formula for the area of a triangle with given sides and circumradius:**
   The area \(A\) of a triangle with sides \(a\), \(b\), and \(c\) and circumradius \(R\) is given by:
   \[
   A = \frac{abc}{4R}
   \]
   Substituting \(a = \sqrt{2} r\), \(b = \sqrt{3} r\), \(c = \sqrt{(2 + \sqrt{3})} r\), and \(R = r\):
   \[
   A = \frac{(\sqrt{2} r)(\sqrt{3} r)(\sqrt{(2 + \sqrt{3})} r)}{4r}
   \]
   Simplifying:
   \[
   A = \frac{\sqrt{6} r^3 \sqrt{(2 + \sqrt{3})}}{4r} = \frac{\sqrt{6} r^2 \sqrt{(2 + \sqrt{3})}}{4}
   \]
   Substituting \(r = \frac{6}{\pi}\):
   \[
   A = \frac{\sqrt{6} \left(\frac{6}{\pi}\right)^2 \sqrt{(2 + \sqrt{3})}}{4} = \frac{\sqrt{6} \left(\frac{36}{\pi^2}\right) \sqrt{(2 + \sqrt{3})}}{4} = \frac{36 \sqrt{6 (2 + \sqrt{3})}}{4 \pi^2}
   \]
   Simplifying further:
   \[
   A = \frac{9 \sqrt{6 (2 + \sqrt{3})}}{\pi^2}
   \]
   Since \(\sqrt{6 (2 + \sqrt{3})} = \sqrt{12 + 6\sqrt{3}} = \sqrt{3} + 3\):
   \[
   A = \frac{9 (\sqrt{3} + 3)}{\pi^2}
   \]

The final answer is \(\boxed{\frac{9}{\pi^2} (\sqrt{3} + 3)}\)