Cosnsider the system of equation
`a_(1)x+b_(1)y+c_(1)z=0, a_(2)x+b_(2)y+c_(2)z=0,`
`a_(3)x+b_(3)y+c_(3)z=0` if `|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0`, then the
system has

Options

A. More than two solutions
B. One trivial and one-non trivial solutions
C. No solution
D. Only trivial solution (0,0,0)

Explanation

To solve the problem, we need to analyze the given system of homogeneous equations and the condition on the determinant of the coefficients.

### Step-by-step Solution:

1. **Understanding the Homogeneous System**:
The given system of equations is:
\[ a_1x + b_1y + c_1z = 0 \]
\[ a_2x + b_2y + c_2z = 0 \]
\[ a_3x + b_3y + c_3z = 0 \]
This is a homogeneous system because all equations are equal to zero.

**Hint**: A homogeneous system always has at least one solution, which is the trivial solution (x = 0, y = 0, z = 0).

2. **Determinant Condition**:
We are given that the determinant of the coefficients of the system is zero:
\[ |(a_1, b_1, c_1), (a_2, b_2, c_2), (a_3, b_3, c_3)| = 0 \]
This means that the rows of the coefficient matrix are linearly dependent.

**Hint**: A determinant of zero indicates that the equations do not represent three independent planes in space.

3. **Implication of Zero Determinant**:
Since the determinant is zero, the system does not have a unique solution. Instead, it implies that there are either no solutions or infinitely many solutions. However, since this is a homogeneous system, it cannot have no solutions.

**Hint**: In a homogeneous system, if the determinant is zero, it guarantees the existence of non-trivial solutions.

4. **Conclusion**:
Therefore, the conclusion is that the system has infinitely many solutions. This is because the equations are dependent, and thus they intersect along a line or a plane in three-dimensional space.

**Final Answer**: The system has infinitely many solutions.