`P ,Q` have position vectors ` vec a& vec b` relative to the origin `' O^(prime)&X , Ya n d vec P Q` internally and externally respectgively in the ratio `2:1` Vector ` vec X Y=` `3/2( vec b- vec a)` b. `4/3( vec a- vec b)` c. `5/6( vec b- vec a)` d. `4/3( vec b- vec a)`

Explanation

To solve the problem, we need to find the vector \( \vec{XY} \) given the position vectors of points \( P \) and \( Q \) relative to the origin \( O \) and the points \( X \) and \( Y \). The points \( P \) and \( Q \) divide the segment \( XY \) in the ratio \( 2:1 \) internally and externally, respectively.

### Step-by-step Solution:

1. **Identify the position vectors**:
- Let the position vector of point \( P \) be \( \vec{a} \).
- Let the position vector of point \( Q \) be \( \vec{b} \).

2. **Find the position vector of point \( X \)**:
- Since \( P \) divides \( XY \) internally in the ratio \( 2:1 \), we can use the section formula:
\[ \vec{OX} = \frac{2\vec{b} + 1\vec{a}}{2 + 1} = \frac{2\vec{b} + \vec{a}}{3} \]

3. **Find the position vector of point \( Y \)**:
- Since \( Q \) divides \( XY \) externally in the ratio \( 2:1 \), we can also use the section formula:
\[ \vec{OY} = \frac{2\vec{b} - 1\vec{a}}{2 - 1} = 2\vec{b} - \vec{a} \]

4. **Calculate the vector \( \vec{XY} \)**:
- Now, we need to find \( \vec{XY} = \vec{OY} - \vec{OX} \):
\[ \vec{XY} = \left(2\vec{b} - \vec{a}\right) - \left(\frac{2\vec{b} + \vec{a}}{3}\right) \]

5. **Simplify the expression**:
- First, express \( \vec{OY} \) and \( \vec{OX} \) with a common denominator:
\[ \vec{XY} = \left(2\vec{b} - \vec{a}\right) - \left(\frac{2\vec{b} + \vec{a}}{3}\right) = \frac{(6\vec{b} - 3\vec{a}) - (2\vec{b} + \vec{a})}{3} \]
- Combine the terms:
\[ \vec{XY} = \frac{6\vec{b} - 3\vec{a} - 2\vec{b} - \vec{a}}{3} = \frac{(6\vec{b} - 2\vec{b}) - (3\vec{a} + \vec{a})}{3} = \frac{4\vec{b} - 4\vec{a}}{3} \]

6. **Final result**:
- Thus, we can express \( \vec{XY} \) as:
\[ \vec{XY} = \frac{4}{3}(\vec{b} - \vec{a}) \]

### Conclusion:
The correct answer is option (d): \( \vec{XY} = \frac{4}{3}(\vec{b} - \vec{a}) \).