If `A(z_(1)),B(z_(2))` and `C(z_(3))`are three points in the Argand plane such that `z_(1)+omegaz_(2)+omega^(2)z_(3)=0`, then

Options

A. A,B, C are collinear triangle
B. `triangleABC` is a right triangle
C. `triangleABC` is an equilateral triangle
D. `triangleABC` is right angled isosceles triangle.

Explanation

To solve the problem, we start with the given equation involving the complex numbers \( z_1, z_2, z_3 \):

### Step 1: Write down the given equation
We have:
\[ z_1 + \omega z_2 + \omega^2 z_3 = 0 \]
where \( \omega \) is a primitive cube root of unity, satisfying \( 1 + \omega + \omega^2 = 0 \).

### Step 2: Substitute \( \omega^2 \)
From the property of cube roots of unity, we can express \( \omega^2 \) in terms of \( \omega \):
\[ \omega^2 = -1 - \omega \]
Substituting this into the equation gives:
\[ z_1 + \omega z_2 - (1 + \omega) z_3 = 0 \]
This simplifies to:
\[ z_1 + \omega z_2 - z_3 - \omega z_3 = 0 \]

### Step 3: Rearranging the equation
Rearranging the terms, we have:
\[ z_1 - z_3 = -\omega z_2 + \omega z_3 \]
This can be rewritten as:
\[ z_1 - z_3 = \omega(z_3 - z_2) \]

### Step 4: Analyze the equation
This equation shows that the vector from \( z_3 \) to \( z_1 \) is equal to \( \omega \) times the vector from \( z_2 \) to \( z_3 \). Since \( \omega = e^{i\pi/3} \), it represents a rotation of \( 60^\circ \) in the complex plane.

### Step 5: Conclude about the triangle
The fact that \( z_1 - z_3 \) is a rotation of \( z_3 - z_2 \) implies that the points \( z_1, z_2, z_3 \) form an equilateral triangle in the Argand plane.

### Final Conclusion
Thus, we conclude that triangle \( ABC \) is an equilateral triangle.