In a triangle AOB, R and Q are the points on the side OB and AB respectively such that 3OR = 2RB and 2AQ = 3QB. Let OQ and AR intersect at the point P (where O is origin).
Q. If the point P divides OQ in the ratio of `mu:1`, then `mu` is :

Options

A. `(2)/(19)`
B. `(2)/(17)`
C. `(2)/(15)`
D. `(10)/(9)`

Answer

To solve the problem, we need to find the ratio \( \mu:1 \) in which point \( P \) divides line segment \( OQ \). We will use the concept of section formula and vector representation.

### Step-by-Step Solution:

1. **Understanding the Triangle and Points**:
- Let \( O \) be the origin, \( A \) and \( B \) be points in the plane.
- Let \( R \) be a point on line segment \( OB \) such that \( 3OR = 2RB \).
- Let \( Q \) be a point on line segment \( AB \) such that \( 2AQ = 3QB \).

2. **Finding the Position Vector of \( R \)**:
- Since \( R \) divides \( OB \) in the ratio \( 2:3 \), we can use the section formula:
\[ \vec{R} = \frac{3\vec{O} + 2\vec{B}}{3 + 2} = \frac{3\vec{0} + 2\vec{B}}{5} = \frac{2\vec{B}}{5} \]

3. **Finding the Position Vector of \( Q \)**:
- Since \( Q \) divides \( AB \) in the ratio \( 3:2 \), we can again use the section formula:
\[ \vec{Q} = \frac{2\vec{A} + 3\vec{B}}{2 + 3} = \frac{2\vec{A} + 3\vec{B}}{5} \]

4. **Finding the Position Vector of \( P \)**:
- Let \( P \) divide \( OQ \) in the ratio \( \mu:1 \). By the section formula:
\[ \vec{P} = \frac{\mu \vec{Q} + 1 \cdot \vec{O}}{\mu + 1} = \frac{\mu \vec{Q}}{\mu + 1} \]
- Substituting \( \vec{Q} \):
\[ \vec{P} = \frac{\mu \left(\frac{2\vec{A} + 3\vec{B}}{5}\right)}{\mu + 1} = \frac{2\mu \vec{A} + 3\mu \vec{B}}{5(\mu + 1)} \]

5. **Finding the Position Vector of \( P \) Using \( AR \)**:
- Now, we also need to express \( P \) in terms of \( A \) and \( R \). Let \( P \) divide \( AR \) in some ratio \( k:1 \):
\[ \vec{P} = \frac{k\vec{R} + 1\vec{A}}{k + 1} \]
- Substituting \( \vec{R} \):
\[ \vec{P} = \frac{k\left(\frac{2\vec{B}}{5}\right) + \vec{A}}{k + 1} = \frac{\frac{2k\vec{B}}{5} + \vec{A}}{k + 1} \]

6. **Equating the Two Expressions for \( \vec{P} \)**:
- We have two expressions for \( \vec{P} \):
\[ \frac{2\mu \vec{A} + 3\mu \vec{B}}{5(\mu + 1)} = \frac{\frac{2k\vec{B}}{5} + \vec{A}}{k + 1} \]
- Cross-multiplying gives:
\[ (2\mu \vec{A} + 3\mu \vec{B})(k + 1) = (2k\vec{B} + 5\vec{A})(\mu + 1) \]

7. **Expanding and Rearranging**:
- Expanding both sides:
\[ 2\mu k \vec{A} + 2\mu \vec{A} + 3\mu k \vec{B} + 3\mu \vec{B} = 2k\mu \vec{B} + 5\mu \vec{A} + 2k\vec{B} + 5\vec{A} \]
- Rearranging gives us a system of equations.

8. **Solving for \( \mu \)**:
- By equating coefficients of \( \vec{A} \) and \( \vec{B} \), we can solve for \( \mu \) and \( k \).
- After simplification, we find:
\[ \mu = \frac{10}{9} \]

### Final Answer:
Thus, the value of \( \mu \) is \( \frac{10}{9} \).