If `a+b+c =0 and a^(2)+b^(2)+c^(2)-ab-bc -ca ne 0, AA a, b, c in R` then the system of equations
`ax+by+cz=0, bx +cy+az=0` and `cx+ay+bz=0` has

Options

A. A unique solution
B. Infinte solutions
C. No solution
D. Exactly two solutions

Answer

To solve the given problem, we need to analyze the system of equations:

1. **Given Equations**:
\[ ax + by + cz = 0 \]
\[ bx + cy + az = 0 \]
\[ cx + ay + bz = 0 \]

2. **Condition**:
We are given that \( a + b + c = 0 \) and \( a^2 + b^2 + c^2 - ab - bc - ca \neq 0 \).

3. **Forming the Coefficient Matrix**:
The system can be represented in matrix form as:
\[ \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

4. **Finding the Determinant**:
We need to find the determinant of the coefficient matrix:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

Using the property of determinants, we can expand this determinant:
\[ D = a(bc - ac) - b(bc - ac) + c(ab - ba) \]
This simplifies to:
\[ D = a^3 + b^3 + c^3 - 3abc \]

5. **Using the Condition \( a + b + c = 0 \)**:
From the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \), and since \( a + b + c = 0 \), we have:
\[ D = 0 \cdot (a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]

6. **Conclusion on the Nature of Solutions**:
Since the determinant \( D = 0 \), the system of equations has either a unique solution or infinitely many solutions. However, since \( a^2 + b^2 + c^2 - ab - ac - bc \neq 0 \), it indicates that the system is consistent and dependent, leading to infinitely many solutions.

Thus, the final answer is that the system of equations has **infinitely many solutions**.