The equation `|(x-a,x-b,x-c),(x-b,x-a,x-c),(x-c,x-b,x-a)|=0` (a,b,c are different) is satisfied by (A) `x=(a+b+c0` (B) `x= 1/3 (a+b+c)` (C) `x=0` (D) none of these

Options

A. `x = 0`
B. `x = a`
C. `x = (1)/(3) (a + b +c)`
D. `x = a + b + c`

Answer

To solve the equation given by the determinant \( |(x-a, x-b, x-c), (x-b, x-a, x-c), (x-c, x-b, x-a)| = 0 \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x-a & x-b & x-c \\ x-b & x-a & x-c \\ x-c & x-b & x-a \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can simplify the determinant by adding the first column to the second and third columns:
\[ D = \begin{vmatrix} x-a & (x-b) + (x-a) & (x-c) + (x-a) \\ x-b & (x-a) + (x-b) & (x-c) + (x-b) \\ x-c & (x-b) + (x-c) & (x-a) + (x-c) \end{vmatrix} \]
This results in:
\[ D = \begin{vmatrix} x-a & 2x - a - b & 2x - a - c \\ x-b & 2x - b - a & 2x - b - c \\ x-c & 2x - c - b & 2x - c - a \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Notice that the first column can be factored out:
\[ D = (3x - (a + b + c)) \cdot \begin{vmatrix} 1 & 1 & 1 \\ x-b & x-a & x-c \\ x-c & x-b & x-a \end{vmatrix} \]

### Step 4: Evaluate the Remaining Determinant
Now we can evaluate the remaining determinant. We can perform row operations on the determinant to simplify it:
\[ D = (3x - (a + b + c)) \cdot \begin{vmatrix} 1 & 1 & 1 \\ x-b & x-a & x-c \\ x-c & x-b & x-a \end{vmatrix} \]

### Step 5: Set the Determinant to Zero
For the determinant to equal zero, we have two cases:
1. \( 3x - (a + b + c) = 0 \)
2. The second determinant equals zero, but since \( a, b, c \) are distinct, this case does not apply.

### Step 6: Solve for x
From the first case:
\[ 3x = a + b + c \implies x = \frac{a + b + c}{3} \]

### Conclusion
Thus, the value of \( x \) that satisfies the equation is:
\[ x = \frac{a + b + c}{3} \]

### Final Answer
The correct option is (B) \( x = \frac{1}{3}(a + b + c) \).