In $\triangle ABC$, points $E$ and $F$ lie on $\overline{AC}, \overline{AB}$, respectively. Denote by $P$ the intersection of $\overline{BE}$ and $\overline{CF}$. Compute the maximum possible area of $\triangle ABC$ if $PB = 14$, $PC = 4$, $PE = 7$, $PF = 2$.

[i]Proposed by Eugene Chen[/i]
1. **Identify the given information and the goal:**
   - We are given a triangle \( \triangle ABC \) with points \( E \) and \( F \) on \( \overline{AC} \) and \( \overline{AB} \) respectively.
   - \( P \) is the intersection of \( \overline{BE} \) and \( \overline{CF} \).
   - We are given the lengths \( PB = 14 \), \( PC = 4 \), \( PE = 7 \), and \( PF = 2 \).
   - We need to compute the maximum possible area of \( \triangle ABC \).

2. **Analyze the given lengths and their implications:**
   - Since \( P \) is the intersection of \( \overline{BE} \) and \( \overline{CF} \), it is a point of concurrency.
   - The given lengths suggest that \( P \) might be the centroid of \( \triangle ABC \). However, we need to verify this.

3. **Verify if \( P \) is the centroid:**
   - The centroid of a triangle divides each median into a ratio of 2:1.
   - Given \( PB = 14 \) and \( PE = 7 \), the ratio \( \frac{PB}{PE} = \frac{14}{7} = 2 \).
   - Given \( PC = 4 \) and \( PF = 2 \), the ratio \( \frac{PC}{PF} = \frac{4}{2} = 2 \).
   - These ratios confirm that \( P \) is indeed the centroid of \( \triangle ABC \).

4. **Determine the area of \( \triangle ABC \):**
   - The centroid divides the triangle into six smaller triangles of equal area.
   - The area of \( \triangle PFB \) is one-sixth of the area of \( \triangle ABC \).
   - To maximize the area of \( \triangle PFB \), we need to maximize the product of the lengths involved and ensure that \( \sin \angle FPB = 1 \), which implies that \( \overline{CF} \) and \( \overline{EB} \) are perpendicular.

5. **Calculate the area of \( \triangle PFB \):**
   - The area of \( \triangle PFB \) can be calculated using the formula for the area of a triangle:
     \[
     \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
     \]
   - Here, the base is \( PF = 2 \) and the height is \( PB = 14 \):
     \[
     \text{Area of } \triangle PFB = \frac{1}{2} \times 2 \times 14 = 14
     \]

6. **Calculate the total area of \( \triangle ABC \):**
   - Since \( \triangle PFB \) is one-sixth of the area of \( \triangle ABC \):
     \[
     \text{Area of } \triangle ABC = 6 \times \text{Area of } \triangle PFB = 6 \times 14 = 84
     \]

The final answer is \( \boxed{84} \).