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 problem, we need to analyze the system of equations given the conditions \( a + b + c = 0 \) and \( a^2 + b^2 + c^2 - ab - bc - ca \neq 0 \).

### Step-by-Step Solution:

1. **Understanding the System of Equations**:
The system of equations is:
\[ \begin{align*} ax + by + cz &= 0 \quad (1) \\ bx + cy + az &= 0 \quad (2) \\ cx + ay + bz &= 0 \quad (3) \end{align*} \]
Since the right-hand side of each equation is zero, this is a homogeneous system of equations.

2. **Properties of Homogeneous Systems**:
A homogeneous system always has at least one solution, which is the trivial solution \( x = 0, y = 0, z = 0 \). It can also have infinitely many solutions or a unique solution depending on the determinant of the coefficients.

3. **Forming the Coefficient Matrix**:
The coefficient matrix \( A \) for the system can be written as:
\[ A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \]

4. **Calculating the Determinant**:
We need to calculate the determinant of matrix \( A \):
\[ \text{det}(A) = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Using the determinant formula for a \( 3 \times 3 \) matrix, we can expand it:
\[ \text{det}(A) = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

5. **Calculating the Minor Determinants**:
Now we calculate each of the \( 2 \times 2 \) determinants:
\[ \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \]
\[ \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \]
\[ \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \]

6. **Putting it All Together**:
Substitute back into the determinant:
\[ \text{det}(A) = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]
Simplifying this gives:
\[ \text{det}(A) = acb - a^3 - b^3 + abc + abc - c^3 \]
Rearranging leads us to:
\[ \text{det}(A) = a^3 + b^3 + c^3 - 3abc \]

7. **Using the Condition**:
We know from the problem statement that \( a + b + c = 0 \). Using the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Since \( a + b + c = 0 \), we have:
\[ \text{det}(A) = 0 \cdot (a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]

8. **Conclusion**:
The determinant being zero indicates that the system has either infinitely many solutions or a unique solution. Given that \( a^2 + b^2 + c^2 - ab - ac - bc \neq 0 \) (as per the problem statement), we conclude that the system has infinitely many solutions.

### Final Answer:
The system of equations has **infinitely many solutions**.