If the system of linear equations
`ax +by+ cz =0`
`cx + ay + bz =0`
` bx + cy +az =0` where a,b,c,` in` R are non-zero and distinct, has a non-zero solution, then:

Options

A. `a +b+c=0`
B. `a,b,c` are in A.P.
C. `1/a, 1/b, 1/c` are in A.P.
D. a,b,c are in G.P.

Explanation

To solve the problem, we need to analyze the given system of linear equations and find the conditions under which it has a non-zero solution. The equations are:

1. \( ax + by + cz = 0 \)
2. \( cx + ay + bz = 0 \)
3. \( bx + cy + az = 0 \)

### Step 1: Formulate the Coefficient Matrix
We can represent the system in matrix form as follows:

\[ \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the Determinant
For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero:

\[ D = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} \]

### Step 3: Calculate the Determinant
We can calculate the determinant \( D \) using the formula for a 3x3 matrix:

\[ D = a \begin{vmatrix} a & b \\ c & a \end{vmatrix} - b \begin{vmatrix} c & b \\ b & a \end{vmatrix} + c \begin{vmatrix} c & a \\ b & c \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \)

2. \( \begin{vmatrix} c & b \\ b & a \end{vmatrix} = ac - b^2 \)

3. \( \begin{vmatrix} c & a \\ b & c \end{vmatrix} = c^2 - ab \)

Substituting these back into the determinant \( D \):

\[ D = a(a^2 - bc) - b(ac - b^2) + c(c^2 - ab) \]

### Step 4: Simplify the Determinant
Expanding this gives:

\[ D = a^3 - abc - (abc - b^3) + c^3 - abc \]

Combining like terms:

\[ D = a^3 + b^3 + c^3 - 3abc \]

### Step 5: Set the Determinant to Zero
For the system to have a non-zero solution, we set the determinant to zero:

\[ a^3 + b^3 + c^3 - 3abc = 0 \]

### Step 6: Use the Identity
Using the identity for the sum of cubes, we know that:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

For this product to be zero, either \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

Since \( a, b, c \) are distinct and non-zero, the condition that must hold is:

\[ a + b + c = 0 \]

### Conclusion
Thus, if the system of linear equations has a non-zero solution, we conclude that:

\[ \boxed{a + b + c = 0} \]