If `a != b != c`, are value of x which satisfies the equation
`|(0,x -a,x -b),(x +a,0,x -c),(x +b,x +c,0)| = 0` is given by

Options

A. `x = 0`
B. `x = c`
C. `x =b`
D. `x =a`

Answer

To solve the equation given by the determinant

\[ \begin{vmatrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{vmatrix} = 0, \]

we will proceed step by step.

### Step 1: Write down the determinant

The determinant we need to evaluate is:

\[ D = \begin{vmatrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the property of determinants, we can expand along the first row:

\[ D = 0 \cdot \begin{vmatrix} 0 & x - c \\ x + c & 0 \end{vmatrix} - (x - a) \cdot \begin{vmatrix} x + a & x - c \\ x + b & 0 \end{vmatrix} + (x - b) \cdot \begin{vmatrix} x + a & 0 \\ x + b & x + c \end{vmatrix} \]

The first term is zero because it is multiplied by zero.

### Step 3: Calculate the remaining 2x2 determinants

1. For the second determinant:

\[ \begin{vmatrix} x + a & x - c \\ x + b & 0 \end{vmatrix} = (x + a) \cdot 0 - (x - c)(x + b) = -(x - c)(x + b) \]

2. For the third determinant:

\[ \begin{vmatrix} x + a & 0 \\ x + b & x + c \end{vmatrix} = (x + a)(x + c) - 0 \cdot (x + b) = (x + a)(x + c) \]

### Step 4: Substitute back into the determinant expression

Now substituting back, we have:

\[ D = -(x - a)(-(x - c)(x + b)) + (x - b)(x + a)(x + c) \]

This simplifies to:

\[ D = (x - a)(x - c)(x + b) + (x - b)(x + a)(x + c) \]

### Step 5: Set the determinant equal to zero

Now we need to solve:

\[ (x - a)(x - c)(x + b) + (x - b)(x + a)(x + c) = 0 \]

### Step 6: Find the values of x

To find the values of \(x\) that satisfy this equation, we can test specific values.

1. **Testing \(x = 0\)**:

\[ D = (0 - a)(0 - c)(0 + b) + (0 - b)(0 + a)(0 + c) \]
\[ = (-a)(-c)(b) + (-b)(a)(c) = abc - abc = 0 \]

Thus, \(x = 0\) satisfies the equation.

### Conclusion

The value of \(x\) which satisfies the equation is:

\[ \boxed{0} \]