If `a ne b ne c`, then one 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. A. 0
B. B. b
C. C. c
D. D. a

Answer

To solve the equation given by the determinant of the matrix

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

we will follow these steps:

### Step 1: Write the Determinant
We start by writing the determinant of the matrix:

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

### Step 2: Expand the Determinant
To find the determinant, we can expand along the first row. The determinant can be calculated as follows:

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

Since the first term is multiplied by 0, we can ignore it. Now we will compute the other two determinants.

### Step 3: Calculate the 2x2 Determinants
1. For the first 2x2 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 second 2x2 determinant:

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

### Step 4: Substitute Back into the Determinant
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 to Zero
We need to set \(D = 0\):

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

### Step 6: Test for Values of x
To find values of \(x\), we can test specific values. Let's try \(x = 0\):

Substituting \(x = 0\):

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

This simplifies to:

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

### Conclusion
Since substituting \(x = 0\) gives us \(D = 0\), we conclude that one value of \(x\) that satisfies the equation is:

\[ \boxed{0}. \]