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 $\boxed{\frac{9}{\pi^2} (\sqrt{3} + 3)}$