In triangle $ABC$, the medians and bisectors corresponding to sides $BC$, $CA$, $AB$ are $m_a$, $m_b$, $m_c$ and $w_a$, $w_b$, $w_c$ respectively. $P=w_a \cap m_b$, $Q=w_b \cap m_c$, $R=w_c \cap m_a$. Denote the areas of triangle $ABC$ and $PQR$ by $F_1$ and $F_2$ respectively. Find the least positive constant $m$ such that $\frac{F_1}{F_2}<m$ holds for any $\triangle{ABC}$.
1. **Understanding the Problem:**
   We need to find the least positive constant \( m \) such that \( \frac{F_1}{F_2} < m \) holds for any triangle \( \triangle ABC \), where \( F_1 \) is the area of \( \triangle ABC \) and \( F_2 \) is the area of \( \triangle PQR \). Here, \( P, Q, R \) are the intersections of the angle bisectors and medians of \( \triangle ABC \).

2. **Areal Coordinates:**
   Using areal coordinates, we can express the coordinates of the points \( P, Q, R \):
   - Any point on \( w_a \) (angle bisector of \( A \)) has coordinates \( \left(\frac{x}{x+b+c}, \frac{b}{x+b+c}, \frac{c}{x+b+c}\right) \), where \( x \) is positive.
   - Any point on \( m_b \) (median from \( B \)) has coordinates \( \left(\frac{1}{2+y}, \frac{y}{2+y}, \frac{1}{2+y}\right) \), where \( y \) is positive.

3. **Intersection Points:**
   - The intersection \( P = w_a \cap m_b \) has coordinates \( \left(\frac{c}{b+2c}, \frac{b}{b+2c}, \frac{c}{b+2c}\right) \).
   - Similarly, \( Q = w_b \cap m_c \) has coordinates \( \left(\frac{a}{c+2a}, \frac{a}{c+2a}, \frac{c}{c+2a}\right) \).
   - \( R = w_c \cap m_a \) has coordinates \( \left(\frac{a}{a+2b}, \frac{b}{a+2b}, \frac{b}{a+2b}\right) \).

4. **Area Calculation:**
   The area ratio \( \frac{F_2}{F_1} \) can be calculated using the determinant of the matrix formed by the coordinates of \( P, Q, R \):
   \[
   \frac{F_2}{F_1} = \left| \begin{matrix} 
   \frac{c}{b+2c} & \frac{b}{b+2c} & \frac{c}{b+2c} \\ 
   \frac{a}{c+2a} & \frac{a}{c+2a} & \frac{c}{c+2a} \\ 
   \frac{a}{a+2b} & \frac{b}{a+2b} & \frac{b}{a+2b} 
   \end{matrix} \right|
   \]

5. **Simplifying the Determinant:**
   Simplifying the determinant, we get:
   \[
   \frac{F_2}{F_1} = \frac{ab^2 + bc^2 + ca^2 - 3abc}{(a+2b)(b+2c)(c+2a)}
   \]

6. **Finding the Maximum:**
   To find the least positive constant \( m \) such that \( \frac{F_1}{F_2} < m \), we need to find the maximum value of \( \frac{F_2}{F_1} \). This involves analyzing the expression:
   \[
   \frac{ab^2 + bc^2 + ca^2 - 3abc}{(a+2b)(b+2c)(c+2a)}
   \]
   By setting \( a = b = x \) and \( c \to 2x \), we can approximate the maximum value:
   \[
   \frac{F_2}{F_1} \approx \frac{x \cdot x^2 + x \cdot (2x)^2 + (2x) \cdot x^2 - 3x \cdot x \cdot 2x}{(x+2x)(x+2 \cdot 2x)(2x+2x)} = \frac{x^3 + 4x^3 + 2x^3 - 6x^3}{3x \cdot 5x \cdot 4x} = \frac{x^3}{60x^3} = \frac{1}{60}
   \]

7. **Conclusion:**
   Therefore, the least positive constant \( m \) such that \( \frac{F_1}{F_2} < m \) for any \( \triangle ABC \) is \( 60 \).

The final answer is \( \boxed{60} \)