Let the triangle $ABC$ have area $1$. The interior bisectors of the angles $\angle BAC,\angle ABC, \angle BCA$ intersect the sides $(BC), (AC), (AB) $ and the circumscribed circle of the respective triangle $ABC$ at the points $L$ and $G, N$ and $F, Q$ and $E$. The lines $EF, FG,GE$ intersect the bisectors $(AL), (CQ) ,(BN)$ respectively at points $P, M, R$. Determine the area of the hexagon $LMNPR$.
1. **Identify the incenter and angle bisectors:**
   Let \( I \) be the incenter of triangle \( ABC \). The angle bisectors of \( \angle BAC \), \( \angle ABC \), and \( \angle BCA \) intersect the sides \( BC \), \( AC \), and \( AB \) at points \( L \), \( N \), and \( Q \) respectively. These bisectors also intersect the circumscribed circle of triangle \( ABC \) at points \( G \), \( F \), and \( E \) respectively.

2. **Use the properties of angle bisectors:**
   By the Angle Bisector Theorem, the angle bisectors divide the opposite sides in the ratio of the adjacent sides. For example, \( \frac{BL}{LC} = \frac{AB}{AC} \).

3. **Midpoints and area relations:**
   Since \( P \) is the midpoint of segment \( AI \), and similarly, \( M \) and \( R \) are midpoints of segments \( CI \) and \( BI \) respectively, we can use the fact that the area of a triangle formed by midpoints of a triangle is one-fourth of the area of the original triangle.

4. **Calculate the area of the hexagon:**
   The hexagon \( LMNPR \) can be divided into smaller regions whose areas can be calculated using the properties of the incenter and the midpoints. Specifically, the area of \( \triangle AIN \) is half the area of \( \triangle ABC \) because \( I \) is the incenter. Similarly, the area of \( \triangle AIP \) is one-fourth the area of \( \triangle ABC \) because \( P \) is the midpoint of \( AI \).

5. **Summing up the areas:**
   By adding the areas of the smaller regions, we find that the area of the hexagon \( LMNPR \) is half the area of the triangle \( ABC \).

\[
\text{Area of hexagon } LMNPR = \frac{1}{2} \times \text{Area of } \triangle ABC = \frac{1}{2} \times 1 = \frac{1}{2}
\]

The final answer is \(\boxed{\frac{1}{2}}\).