If `|(x+y,y+z,z+x),(y+z, z+x,x+y),(z+x,x+y,y+z)|`=k`|(x,z,y),(y,x,z),(z,y,x)|`, then k is equal to

Options

A. `-2`
B. 2
C. `-3`
D. 3

Answer

To solve the given problem, we need to evaluate the determinants and find the value of \( k \) such that:

\[ |(x+y, y+z, z+x), (y+z, z+x, x+y), (z+x, x+y, y+z)| = k |(x, z, y), (y, x, z), (z, y, x)| \]

Let's denote the first determinant as \( D_1 \) and the second determinant as \( D_2 \).

### Step 1: Evaluate \( D_1 \)

We have:

\[ D_1 = \begin{vmatrix} x+y & y+z & z+x \\ y+z & z+x & x+y \\ z+x & x+y & y+z \end{vmatrix} \]

To simplify \( D_1 \), we can use the property of determinants that allows us to add or subtract multiples of rows or columns.

1. **Add all columns**:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ D_1 = \begin{vmatrix} 2(x+y+z) & y+z & z+x \\ 2(x+y+z) & z+x & x+y \\ 2(x+y+z) & x+y & y+z \end{vmatrix} \]

2. **Factor out \( 2(x+y+z) \)** from the first column:
\[ D_1 = 2(x+y+z) \begin{vmatrix} 1 & y+z & z+x \\ 1 & z+x & x+y \\ 1 & x+y & y+z \end{vmatrix} \]

3. **Subtract the first row from the second and third rows**:
\[ R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1 \]
This gives us:
\[ D_1 = 2(x+y+z) \begin{vmatrix} 1 & y+z & z+x \\ 0 & (z+x)-(y+z) & (x+y)-(y+z) \\ 0 & (x+y)-(y+z) & (y+z)-(z+x) \end{vmatrix} \]

4. **Simplifying further**:
\[ D_1 = 2(x+y+z) \begin{vmatrix} 1 & y+z & z+x \\ 0 & x-y & x-z \\ 0 & y-z & y-x \end{vmatrix} \]

5. **The determinant of the \( 2 \times 2 \) matrix** can be computed:
\[ D_1 = 2(x+y+z) \cdot (x-y)(y-z)(z-x) \]

### Step 2: Evaluate \( D_2 \)

Now, we evaluate \( D_2 \):

\[ D_2 = \begin{vmatrix} x & z & y \\ y & x & z \\ z & y & x \end{vmatrix} \]

This determinant can be computed directly or using properties of determinants.

1. **Using the property of determinants**:
\[ D_2 = (x-y)(y-z)(z-x) \]

### Step 3: Relate \( D_1 \) and \( D_2 \)

Now we have:

\[ D_1 = 2(x+y+z)(x-y)(y-z)(z-x) \]
\[ D_2 = (x-y)(y-z)(z-x) \]

We can set up the equation:

\[ 2(x+y+z)(x-y)(y-z)(z-x) = k(x-y)(y-z)(z-x) \]

### Step 4: Solve for \( k \)

Dividing both sides by \( (x-y)(y-z)(z-x) \) (assuming \( x, y, z \) are distinct):

\[ 2(x+y+z) = k \]

Thus, we find:

\[ k = 2(x+y+z) \]

### Final Answer

The value of \( k \) is:

\[ \boxed{2} \]