If a, b and c are real numbers, then the value of the determinant `|(1-a,a-b-c,b+c),(1-b,b-c-a,c+a),(1-c,c-a-b,a+b)|` is

Options

A. A) 0
B. B) (a - b) (b - c) (c - a)
C. C) `(a + b + c)^(2)`
D. D) `(a + b + c)^(3)`

Answer

To find the value of the determinant

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

we will perform a series of column operations to simplify the determinant.

### Step 1: Column Operation
We will first perform the operation \( C_2 \leftarrow C_2 + C_3 \). This means we will add the third column to the second column.

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

Calculating the new elements in \( C_2 \):

- For the first row: \( (a-b-c) + (b+c) = a \)
- For the second row: \( (b-c-a) + (c+a) = b \)
- For the third row: \( (c-a-b) + (a+b) = c \)

Thus, we have:

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

### Step 2: Another Column Operation
Next, we will perform the operation \( C_1 \leftarrow C_1 + C_2 \).

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

Calculating the new elements in \( C_1 \):

- For the first row: \( (1-a) + a = 1 \)
- For the second row: \( (1-b) + b = 1 \)
- For the third row: \( (1-c) + c = 1 \)

Thus, we have:

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

### Step 3: Identical Rows
Now we notice that the first column has identical entries (all are 1). According to the properties of determinants, if two rows (or columns) are identical, the value of the determinant is zero.

Thus, we conclude that:

\[ D = 0 \]

### Final Answer
The value of the determinant is \( \boxed{0} \).

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