In $\triangle{ABC}$ with side lengths $AB = 13$, $AC = 12$, and $BC = 5$, let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$. What is the area of $\triangle{MOI}$?

$\textbf{(A)}\ 5/2\qquad\textbf{(B)}\ 11/4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13/4\qquad\textbf{(E)}\ 7/2$
1. **Place the triangle on the coordinate plane:**
   - Let \( C = (0,0) \), \( A = (0,12) \), and \( B = (5,0) \).
   - Since \(\triangle ABC\) is a right triangle, the circumcenter \( O \) is the midpoint of the hypotenuse \( AB \).
   - The coordinates of \( O \) are:
     \[
     O = \left( \frac{0+5}{2}, \frac{12+0}{2} \right) = \left( \frac{5}{2}, 6 \right)
     \]

2. **Find the incenter \( I \):**
   - The area of \(\triangle ABC\) is:
     \[
     \text{Area} = \frac{1}{2} \cdot 5 \cdot 12 = 30
     \]
   - The semi-perimeter \( s \) is:
     \[
     s = \frac{13 + 12 + 5}{2} = 15
     \]
   - The inradius \( r \) is:
     \[
     r = \frac{\text{Area}}{s} = \frac{30}{15} = 2
     \]
   - The incenter \( I \) lies on the angle bisector and is equidistant from \( AC \) and \( BC \). Since \( I \) is 2 units away from both axes, its coordinates are:
     \[
     I = (2, 2)
     \]

3. **Find the coordinates of \( M \):**
   - Let \( M = (x, x) \) since it is tangent to both \( AC \) and \( BC \).
   - The radius of the circle centered at \( M \) is \( x \).
   - The circle is also tangent to the circumcircle of \(\triangle ABC\). The circumradius \( R \) of \(\triangle ABC\) is:
     \[
     R = \frac{13}{2}
     \]
   - The distance \( OM \) plus the radius \( x \) equals the circumradius \( R \):
     \[
     OM + x = \frac{13}{2}
     \]
   - Using the distance formula for \( OM \):
     \[
     OM = \sqrt{\left( \frac{5}{2} - x \right)^2 + (6 - x)^2}
     \]
   - Therefore:
     \[
     \sqrt{\left( \frac{5}{2} - x \right)^2 + (6 - x)^2} + x = \frac{13}{2}
     \]
   - Squaring both sides:
     \[
     \left( \frac{5}{2} - x \right)^2 + (6 - x)^2 = \left( \frac{13}{2} - x \right)^2
     \]
   - Expanding and simplifying:
     \[
     \left( \frac{25}{4} - 5x + x^2 \right) + (36 - 12x + x^2) = \left( \frac{169}{4} - 13x + x^2 \right)
     \]
     \[
     \frac{25}{4} + 36 - 5x - 12x + 2x^2 = \frac{169}{4} - 13x + x^2
     \]
     \[
     \frac{169}{4} - 17x + 2x^2 = \frac{169}{4} - 13x + x^2
     \]
     \[
     2x^2 - 17x = x^2 - 13x
     \]
     \[
     x^2 - 4x = 0
     \]
     \[
     x(x - 4) = 0
     \]
     \[
     x = 4 \quad (\text{since } x \neq 0)
     \]
   - Thus, the coordinates of \( M \) are:
     \[
     M = (4, 4)
     \]

4. **Calculate the area of \(\triangle MOI\) using the Shoelace formula:**
   - The vertices are \( O = \left( \frac{5}{2}, 6 \right) \), \( I = (2, 2) \), and \( M = (4, 4) \).
   - Using the Shoelace formula:
     \[
     \text{Area} = \frac{1}{2} \left| \frac{5}{2} \cdot 2 + 2 \cdot 4 + 4 \cdot 6 - (6 \cdot 2 + 2 \cdot 4 + 4 \cdot \frac{5}{2}) \right|
     \]
     \[
     = \frac{1}{2} \left| 5 + 8 + 24 - (12 + 8 + 10) \right|
     \]
     \[
     = \frac{1}{2} \left| 37 - 30 \right|
     \]
     \[
     = \frac{1}{2} \cdot 7 = \frac{7}{2}
     \]

The final answer is \(\boxed{\frac{7}{2}}\)