Let `z_(1),z_(2),z_(3)` be the three nonzero comple numbers such that `z_(1) ne 1, a= |z_(1)|, b = |z_(2)| and c= |z_(3)|`. Let `|{:(,a,b,c),(,b,c,a),(,c,a,b):}|= 0` Then

Options

A. `arg((z_(3))/(z_(2))) = arg((z_(3)-z_(1))/(z_(2)-z_(1)))`
B. ortho centre of triangle formed by `z_(1),z_(2),z_(3)` is `z_(1)+z_(2)+z_(3)`
C. if trinagle formed by `z_(1),z_(2),z_(3) ` is `z_(1)+z_(2)+z_(3)` is `(3sqrt(3))/(2)|z_(1)|^(2)`
D. if triangle formed by `z_(1),z_(2),z_(3)` is equlateral, then `z_(1)+z_(2)+z_(3)=0`

Explanation

To solve the problem, we need to analyze the given determinant condition and derive the necessary relationships between the moduli of the complex numbers \( z_1, z_2, z_3 \).

### Step-by-Step Solution:

1. **Given Condition**:
We are given that the determinant of the following matrix is zero:
\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]
where \( a = |z_1| \), \( b = |z_2| \), and \( c = |z_3| \).

2. **Expanding the Determinant**:
We can expand the determinant using the formula for a \( 3 \times 3 \) determinant:
\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]
This gives us:
\[ D = a(cb - a^2) - b(bc - ac) + c(ab - c^2) \]

3. **Simplifying the Expression**:
Expanding the terms:
\[ D = acb - a^3 - b^2c + abc + abc - c^3 \]
Combining like terms:
\[ D = 3abc - (a^3 + b^3 + c^3) \]

4. **Setting the Determinant to Zero**:
Since the determinant is given to be zero, we have:
\[ 3abc - (a^3 + b^3 + c^3) = 0 \]
Rearranging gives:
\[ a^3 + b^3 + c^3 = 3abc \]

5. **Using the Identity**:
The identity \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \) implies that either:
- \( a + b + c = 0 \) (not possible since \( a, b, c > 0 \)), or
- \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

6. **Conclusion**:
The second condition implies:
\[ (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \]
Therefore, \( a = b = c \).

### Final Result:
Since \( a = |z_1| \), \( b = |z_2| \), and \( c = |z_3| \), we conclude that:
\[ |z_1| = |z_2| = |z_3| \]