Let $ABC$ be a triangle with sides $51, 52, 53$. Let $\Omega$ denote the incircle of $\bigtriangleup ABC$. Draw tangents to $\Omega$ which are parallel to the sides of $ABC$. Let $r_1, r_2, r_3$ be the inradii of the three corener triangles so formed, Find the largest integer that does not exceed $r_1 + r_2 + r_3$.
1. **Calculate the inradius \( r \) of \(\triangle ABC\):**
   The formula for the inradius \( r \) of a triangle is given by:
   \[
   r = \frac{\Delta}{s}
   \]
   where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter. First, we need to find \(\Delta\) and \(s\).

2. **Calculate the semi-perimeter \( s \):**
   \[
   s = \frac{a + b + c}{2} = \frac{51 + 52 + 53}{2} = 78
   \]

3. **Calculate the area \(\Delta\) using Heron's formula:**
   Heron's formula states:
   \[
   \Delta = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   Substituting the values:
   \[
   \Delta = \sqrt{78(78-51)(78-52)(78-53)} = \sqrt{78 \cdot 27 \cdot 26 \cdot 25}
   \]
   Simplifying inside the square root:
   \[
   \Delta = \sqrt{78 \cdot 27 \cdot 26 \cdot 25} = \sqrt{78 \cdot 27 \cdot 650} = \sqrt{1377000}
   \]
   \[
   \Delta = 1050
   \]

4. **Calculate the inradius \( r \):**
   \[
   r = \frac{\Delta}{s} = \frac{1050}{78} = 15
   \]

5. **Determine the scaling factor for the smaller triangles:**
   The triangles formed by the tangents parallel to the sides of \(\triangle ABC\) are similar to \(\triangle ABC\). Let \(T_a\), \(T_b\), and \(T_c\) be the smaller triangles near vertices \(A\), \(B\), and \(C\) respectively. The scaling factor is the ratio of the heights of these triangles to the height of \(\triangle ABC\).

6. **Express the inradii \(r_1, r_2, r_3\) in terms of \(r\):**
   The height of \(\triangle ABC\) from vertex \(A\) is \(H_a = 2r + h_a = 30 + h_a\). The inradius \(r_a\) of \(T_a\) is:
   \[
   r_a = r \cdot \frac{h_a}{H_a} = r \cdot \frac{H_a - 30}{H_a}
   \]
   Similarly for \(r_b\) and \(r_c\):
   \[
   r_b = r \cdot \frac{H_b - 30}{H_b}, \quad r_c = r \cdot \frac{H_c - 30}{H_c}
   \]

7. **Sum the inradii \(r_1, r_2, r_3\):**
   \[
   \sum r_i = r \left( 3 - 30 \left( \sum \frac{1}{H_i}\right)\right)
   \]
   We need to compute \(\sum \frac{1}{H_i}\).

8. **Calculate \(\sum \frac{1}{H_i}\):**
   \[
   \sum \frac{1}{H_i} = \frac{1}{2} \left(\frac{a}{\Delta} + \frac{b}{\Delta} + \frac{c}{\Delta} \right) = \frac{s}{\Delta} = \frac{78}{1050} = \frac{1}{15}
   \]

9. **Substitute back to find \(\sum r_i\):**
   \[
   \sum r_i = 15\left(3-30\cdot \frac{1}{15}\right) = 15 \left(3 - 2\right) = 15 \cdot 1 = 15
   \]

The final answer is \(\boxed{15}\).