` " if " |{:(b+c,,c+a,,a+b),(a+b,,b+c,,c+a),(c+a,,a+b,,b+c):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|` then the value of k is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem, we need to evaluate the determinants and find the value of \( k \). Let's break down the solution step by step.

### Step 1: Write down the determinants
We have two determinants to consider:

1. \( D_1 = \begin{vmatrix} b+c & c+a & a+b \\ a+b & b+c & c+a \\ c+a & a+b & b+c \end{vmatrix} \)

2. \( D_2 = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} \)

We need to find \( k \) such that \( D_1 = k \cdot D_2 \).

### Step 2: Simplify the first determinant \( D_1 \)
We can apply properties of determinants to simplify \( D_1 \).

**Operation:** Replace the first column \( C_1 \) with \( C_1 + C_2 + C_3 \).

This gives us:

\[ D_1 = \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & b+c & c+a \\ 2(a+b+c) & a+b & b+c \end{vmatrix} \]

### Step 3: Factor out the common term
Since \( 2(a+b+c) \) is common in the first column, we can factor it out:

\[ D_1 = 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 1 & b+c & c+a \\ 1 & a+b & b+c \end{vmatrix} \]

### Step 4: Further simplify the determinant
Now, we can perform row operations on the determinant:

**Operation:** Replace \( R_2 \) with \( R_2 - R_1 \) and \( R_3 \) with \( R_3 - R_1 \):

\[ D_1 = 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 0 & (b+c) - (c+a) & (a+b) - (c+a) \\ 0 & (a+b) - (c+a) & (b+c) - (c+a) \end{vmatrix} \]

This simplifies to:

\[ D_1 = 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 0 & b-a & b-c \\ 0 & a-b & a-c \end{vmatrix} \]

### Step 5: Calculate the determinant
The determinant of a matrix with a row of zeros can be calculated as follows:

\[ D_1 = 2(a+b+c) \cdot 1 \cdot \begin{vmatrix} b-a & b-c \\ a-b & a-c \end{vmatrix} \]

Calculating this determinant gives:

\[ D_1 = 2(a+b+c) \cdot ((b-a)(a-c) - (b-c)(a-b)) \]

### Step 6: Evaluate the second determinant \( D_2 \)
Now, we calculate \( D_2 \):

\[ D_2 = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} \]

Using the properties of determinants, we can find that:

\[ D_2 = (a-b)(b-c)(c-a) \]

### Step 7: Set up the equation
Now we have:

\[ D_1 = k \cdot D_2 \]

Substituting the values we found:

\[ 2(a+b+c) \cdot ((b-a)(a-c) - (b-c)(a-b)) = k \cdot (a-b)(b-c)(c-a) \]

### Step 8: Solve for \( k \)
By comparing coefficients, we can find that \( k = 2 \).

### Conclusion
Thus, the value of \( k \) is:

\[ \boxed{2} \]