Let `A(z_(1)),B(z_(2)) and C(z_(3))` be complex numbers satisfying the equation`|z|=1` and also satisfying the relation `3z_(1)=2z_(2)+2z_(3)`. Then `|z_(2)-z_(3)|^(2)` is equal to

Answer

To solve the problem, we need to find \( |z_2 - z_3|^2 \) given the conditions that \( |z| = 1 \) and \( 3z_1 = 2z_2 + 2z_3 \).

### Step-by-step solution:

1. **Understanding the given conditions**:
- We know that \( |z_1| = |z_2| = |z_3| = 1 \). This means that the complex numbers \( z_1, z_2, z_3 \) lie on the unit circle in the complex plane.

2. **Rearranging the equation**:
- From the equation \( 3z_1 = 2z_2 + 2z_3 \), we can rearrange it to find a relationship between \( z_2 \) and \( z_3 \):
\[ 3z_1 = 2(z_2 + z_3) \implies z_2 + z_3 = \frac{3}{2} z_1 \]

3. **Finding the midpoint**:
- The equation \( z_2 + z_3 = \frac{3}{2} z_1 \) suggests that \( z_2 \) and \( z_3 \) are symmetrically located around the point \( \frac{3}{4} z_1 \) on the unit circle. We can express \( z_2 \) and \( z_3 \) in terms of their midpoint \( p \):
\[ z_2 + z_3 = 2p \quad \text{where } p = \frac{3}{4} z_1 \]

4. **Expressing \( z_2 \) and \( z_3 \)**:
- Let \( z_2 = p + d \) and \( z_3 = p - d \) for some complex number \( d \). Then:
\[ z_2 - z_3 = (p + d) - (p - d) = 2d \]

5. **Calculating \( |z_2 - z_3|^2 \)**:
- We need to find \( |z_2 - z_3|^2 \):
\[ |z_2 - z_3|^2 = |2d|^2 = 4|d|^2 \]

6. **Finding \( |d| \)**:
- Since \( z_1, z_2, z_3 \) lie on the unit circle, we can use the Pythagorean theorem to find \( |d| \). The distance from the origin to \( p \) is:
\[ |p| = \left|\frac{3}{4} z_1\right| = \frac{3}{4} \]
- The distance from the origin to \( z_2 \) and \( z_3 \) is 1. Thus, we can set up the equation:
\[ |z_2|^2 = |p + d|^2 = 1 \quad \text{and} \quad |z_3|^2 = |p - d|^2 = 1 \]

7. **Using the distance formula**:
- Expanding \( |p + d|^2 \):
\[ |p + d|^2 = |p|^2 + |d|^2 + 2 \text{Re}(p \overline{d}) = 1 \]
- Since \( |p|^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \):
\[ \frac{9}{16} + |d|^2 + 2 \text{Re}(p \overline{d}) = 1 \]
- This leads to:
\[ |d|^2 + 2 \text{Re}(p \overline{d}) = 1 - \frac{9}{16} = \frac{7}{16} \]

8. **Finding \( |d|^2 \)**:
- Since \( p \) is a point on the unit circle, and \( d \) is the distance from the midpoint to either \( z_2 \) or \( z_3 \), we can conclude that \( |d|^2 = \frac{7}{16} \).

9. **Final calculation**:
- Thus, substituting back:
\[ |z_2 - z_3|^2 = 4|d|^2 = 4 \cdot \frac{7}{16} = \frac{7}{4} \]

### Conclusion:
The value of \( |z_2 - z_3|^2 \) is \( \frac{7}{4} \).