The value of `|{:(1,1,1),(b+c,c+a,a+b),(b+c-a,c+a-b,a+b-c):}|` is

Options

A. `a+b+c`
B. `2(a+b+c)`
C. 1
D. 0

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{vmatrix} \]

we will perform some row operations to simplify it.

### Step 1: Write the determinant
We start by writing the determinant:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{vmatrix} \]

### Step 2: Perform column operations
We will perform column operations to simplify the determinant. Specifically, we will subtract the first column from the second column and the second column from the third column.

\[ D = \begin{vmatrix} 1 & 1-1 & 1-1 \\ b+c & (c+a) - (b+c) & (a+b) - (c+a) \\ b+c-a & (c+a-b) - (b+c-a) & (a+b-c) - (c+a-b) \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} 1 & 0 & 0 \\ b+c & a-b & b-c \\ b+c-a & a-b & 2a-b-c \end{vmatrix} \]

### Step 3: Simplify further
Now we can see that the first column is \(1, b+c, b+c-a\). We can factor out common terms from the second and third columns.

\[ D = \begin{vmatrix} 1 & 0 & 0 \\ b+c & a-b & b-c \\ b+c-a & a-b & 2a-b-c \end{vmatrix} \]

### Step 4: Expand the determinant
Now we can expand the determinant along the first row:

\[ D = 1 \cdot \begin{vmatrix} a-b & b-c \\ a-b & 2a-b-c \end{vmatrix} \]

### Step 5: Calculate the 2x2 determinant
Calculating the 2x2 determinant:

\[ \begin{vmatrix} a-b & b-c \\ a-b & 2a-b-c \end{vmatrix} = (a-b)(2a-b-c) - (a-b)(b-c) \]

Factoring out \((a-b)\):

\[ = (a-b) \left( (2a-b-c) - (b-c) \right) = (a-b)(2a - 2b) = (a-b)(2(a-b)) \]

### Step 6: Final result
Thus, we have:

\[ D = (a-b)(2(a-b)) = 2(a-b)^2 \]

### Conclusion
The final value of the determinant is \(0\) when \(a = b\) or \(a = c\) or \(b = c\). Therefore, the answer is:

\[ \text{The value of the determinant is } 0. \]