, a point 'z' is equidistant from three distinct points `z_(1),z_(2)` and `z_(3)` in the Argand plane. If `z,z_(1)` and `z_(2)` are collinear, then arg`(z(z_(3)-z_(1))/(z_(3)-z_(2))). Will be `(z_(1),z_(2),z_(3))` are in anticlockwise sense).

Options

A. `pi/2`
B. `-pi/2`
C. `pi/6`
D. `(2pi)/3`

Explanation

To solve the problem step-by-step, we will analyze the conditions given and apply properties of complex numbers and their arguments.

### Step 1: Understand the Given Information
We have three distinct points \( z_1, z_2, z_3 \) in the Argand plane, and a point \( z \) that is equidistant from these three points. The point \( z \) lies on the line segment joining \( z_1 \) and \( z_2 \), meaning \( z \) is on the line defined by \( z_1 \) and \( z_2 \).

### Step 2: Use the Equidistance Condition
Since \( z \) is equidistant from \( z_1, z_2, \) and \( z_3 \), we can write:
\[ |z - z_1| = |z - z_2| = |z - z_3| \]
This implies that \( z \) lies on the perpendicular bisector of the segment joining \( z_1 \) and \( z_2 \).

### Step 3: Establish the Argument Condition
Given that \( z, z_1, \) and \( z_2 \) are collinear, we can express \( z \) in terms of \( z_1 \) and \( z_2 \). Let’s denote the point \( z \) as \( z = \lambda z_1 + (1 - \lambda) z_2 \) for some \( \lambda \in [0, 1] \).

### Step 4: Analyze the Argument Expression
We need to find:
\[ \arg\left(\frac{z_3 - z_1}{z_3 - z_2}\right) \]
Using the properties of arguments, we can express this as:
\[ \arg(z_3 - z_1) - \arg(z_3 - z_2) \]

### Step 5: Geometric Interpretation
Since \( z_1, z_2, z_3 \) are points in the Argand plane, the angles formed by these points can be analyzed. The condition that \( z_1, z_2, z_3 \) are in anticlockwise sense means that the angle from \( z_2 \) to \( z_3 \) to \( z_1 \) is positive.

### Step 6: Use the Properties of Angles
From the properties of angles in a triangle, we know:
- If \( z_1, z_2, z_3 \) are in anticlockwise order, then:
\[ \arg(z_3 - z_1) - \arg(z_3 - z_2) = \text{angle } ACB \]
where \( A = z_1, B = z_2, C = z_3 \).

### Step 7: Conclusion
Since \( z_1, z_2, z_3 \) are in anticlockwise order, the angle \( ACB \) is positive. Thus, we conclude:
\[ \arg\left(\frac{z_3 - z_1}{z_3 - z_2}\right) = \theta \]
where \( \theta \) is the angle formed, which is positive.

### Final Result
The argument \( \arg\left(\frac{z_3 - z_1}{z_3 - z_2}\right) \) is equal to \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \) depending on the orientation of the points, but since \( z_1, z_2, z_3 \) are anticlockwise, we conclude:
\[ \arg\left(\frac{z_3 - z_1}{z_3 - z_2}\right) = \frac{\pi}{2} \]