Line `OQ` is angle bisector of angle `O` of right angle triangle `OPR`, right angle at `P`. Point `Q` is such that `ORQP` is concyclic. If point `O` is orign and points `P,Q,R` are represented by the complex numbers `z_(3),z_(2),z_(1)` respectively. If `(z_(2)^(2))/(z_(1)z_(2))=3/2` then (`R` is circum radius of `/_\OPR`)

Options

A. Angles of `/_\OPR` are `(pi)/6,(pi)/3,(pi)/2`
B. Angles of `/_\OPR` are `(pi)/4,(pi)/4,(pi)/2`
C. Area of `/_\OPR` is `2sqrt(2)R^(2)`
D. Area of `/_\OPR` is `2R^(2)`

Explanation

To solve the problem step by step, we will analyze the given information and use the properties of triangles and complex numbers.

### Step 1: Understand the Geometry
We have a right triangle \( OPR \) with \( O \) as the origin, \( P \) at \( z_3 \), \( Q \) at \( z_2 \), and \( R \) at \( z_1 \). The angle bisector \( OQ \) divides angle \( O \) into two equal angles. The points \( O, R, Q, P \) are concyclic.

**Hint:** Draw the triangle \( OPR \) and mark the points \( O, P, R \) and the angle bisector \( OQ \).

### Step 2: Use the Given Condition
We are given that:
\[ \frac{z_2^2}{z_1 z_3} = \frac{3}{2} \]
This can be rearranged to:
\[ z_2^2 = \frac{3}{2} z_1 z_3 \]

**Hint:** This equation relates the complex numbers representing the points in the triangle.

### Step 3: Apply Rotation Theory
Using rotation theory, we can express the relationships between the complex numbers:
\[ \frac{z_2}{z_1} = \cos \theta e^{i\theta} \]
\[ \frac{z_3}{z_2} = \cos 2\theta e^{i\theta} \]

**Hint:** This uses the properties of angles and rotations in the complex plane.

### Step 4: Combine the Relationships
From the above expressions, we can write:
\[ \frac{z_2}{z_1} \cdot \frac{z_2}{z_3} = \frac{z_2^2}{z_1 z_3} = \cos \theta e^{i\theta} \cdot \frac{1}{\cos 2\theta e^{i\theta} \cos \theta} \]
This simplifies to:
\[ \frac{z_2^2}{z_1 z_3} = \frac{\cos^2 \theta}{\cos 2\theta} \]
Setting this equal to \( \frac{3}{2} \):
\[ \frac{\cos^2 \theta}{\cos 2\theta} = \frac{3}{2} \]

**Hint:** This step involves substituting the relationships into the given condition.

### Step 5: Use the Double Angle Formula
Using the double angle formula:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Substituting this into our equation gives:
\[ \frac{\cos^2 \theta}{2\cos^2 \theta - 1} = \frac{3}{2} \]

**Hint:** Remember to simplify the equation correctly.

### Step 6: Solve for \( \cos \theta \)
Cross-multiplying gives:
\[ 2\cos^2 \theta = 3(2\cos^2 \theta - 1) \]
This simplifies to:
\[ 2\cos^2 \theta = 6\cos^2 \theta - 3 \]
Rearranging gives:
\[ 4\cos^2 \theta = 3 \implies \cos^2 \theta = \frac{3}{4} \implies \cos \theta = \frac{\sqrt{3}}{2} \]
Thus, \( \theta = \frac{\pi}{6} \).

**Hint:** Recognize that \( \theta \) corresponds to one of the angles in the triangle.

### Step 7: Find the Angles of Triangle \( OPR \)
Since \( OPR \) is a right triangle at \( P \):
- One angle \( O = \theta = \frac{\pi}{6} \)
- The other angle \( R = \frac{\pi}{2} \)
- The third angle \( P = \pi - \left(\frac{\pi}{6} + \frac{\pi}{2}\right) = \frac{\pi}{3} \)

Thus, the angles of triangle \( OPR \) are \( \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2} \).

**Hint:** Use the property that the sum of angles in a triangle is \( \pi \).

### Step 8: Calculate the Area of Triangle \( OPR \)
The area of triangle \( OPR \) can be calculated using:
\[ \text{Area} = \frac{1}{2} \times OP \times PR \]
Given that \( OP = 2R \) and \( PR = 2R \) (since \( OQRP \) is cyclic), we have:
\[ \text{Area} = \frac{1}{2} \times (2R) \times (2R) = 2R^2 \]

**Hint:** Remember the formula for the area of a right triangle.

### Final Answer
The angles of triangle \( OPR \) are \( \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2} \) and the area is \( 2R^2 \).