In a parallelogram `OABC,` vectors `vec a, vec b, vec c` are respectively the positions of vectors of vertices `A, B, C` with reference to O as origin. A point E is taken on the side BC which divide the line `2:1` internally. Also the line segment AE intersect the line bisecting the angle O internally in point P. If CP, when extended meets AB in point F. Then The position vector of point P, is

Explanation

To solve the problem step by step, we will follow the given information about the parallelogram \( OABC \) and the points defined in the question.

### Step 1: Understand the Geometry
In the parallelogram \( OABC \):
- The position vectors of points \( A, B, C \) are given as \( \vec{a}, \vec{b}, \vec{c} \) respectively.
- \( O \) is the origin.

### Step 2: Find the Position Vector of Point \( E \)
Point \( E \) divides the line segment \( BC \) in the ratio \( 2:1 \) internally. Using the section formula, the position vector of point \( E \) can be calculated as follows:
\[ \vec{E} = \frac{2\vec{c} + 1\vec{b}}{2 + 1} = \frac{2\vec{c} + \vec{b}}{3} \]

### Step 3: Find the Position Vector of Point \( P \)
Point \( P \) lies on the angle bisector of \( \angle AOB \). The position vector of \( P \) can be expressed in terms of \( \vec{a} \) and \( \vec{b} \) as:
\[ \vec{P} = \lambda \vec{a} + \mu \vec{b} \]
where \( \lambda \) and \( \mu \) are proportional to the magnitudes of \( \vec{b} \) and \( \vec{a} \) respectively.

### Step 4: Use the Angle Bisector Theorem
According to the angle bisector theorem, we can express \( \vec{P} \) as:
\[ \vec{P} = \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \]

### Step 5: Find the Ratio in Which \( P \) Divides \( E \)
Since \( P \) also divides \( E \) in the ratio \( 2:1 \), we can express this as:
\[ \vec{P} = \frac{2\vec{E} + 1\vec{A}}{2 + 1} = \frac{2\left(\frac{2\vec{c} + \vec{b}}{3}\right) + \vec{a}}{3} \]
This simplifies to:
\[ \vec{P} = \frac{\frac{4\vec{c} + 2\vec{b}}{3} + \vec{a}}{3} = \frac{4\vec{c} + 2\vec{b} + 3\vec{a}}{9} \]

### Step 6: Equate the Two Expressions for \( \vec{P} \)
Now we have two expressions for \( \vec{P} \):
1. From the angle bisector: \( \vec{P} = \lambda \vec{a} + \mu \vec{b} \)
2. From the section formula: \( \vec{P} = \frac{4\vec{c} + 2\vec{b} + 3\vec{a}}{9} \)

### Step 7: Solve for \( \lambda \) and \( \mu \)
By comparing coefficients from both expressions, we can solve for \( \lambda \) and \( \mu \).

### Final Step: Conclusion
After solving, we can conclude that the position vector of point \( P \) is given by:
\[ \vec{P} = \frac{4\vec{c} + 2\vec{b} + 3\vec{a}}{9} \]