If `a + x = b + y = c + z + 1` , where a, b,c,x,y,z are non polar distinct real numbers , then `|(x, a+y,x + a),(y , b+y , y + b),(z,c+y, z+c)|` is equal to :

Options

A. `0 `
B. `y(a-b) `
C. `y(b -c) `
D. `y(a-c) `

Answer

To solve the problem, we need to find the determinant of the matrix given the conditions provided. Let's denote the determinant as \( D \).

Given:
\[ D = \begin{vmatrix} x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c \end{vmatrix} \]

### Step 1: Simplifying the Determinant
First, we will perform column operations to simplify the determinant. We will subtract the first column from the third column.

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

This simplifies to:
\[ D = \begin{vmatrix} x & a+y & a \\ y & b+y & b \\ z & c+y & c \end{vmatrix} \]

### Step 2: Further Simplification
Next, we will perform another column operation by subtracting the second column from the third column.

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

This simplifies to:
\[ D = \begin{vmatrix} x & a+y & -y \\ y & b+y & -y \\ z & c+y & -y \end{vmatrix} \]

### Step 3: Factoring Out Common Terms
Now, we can factor out \(-y\) from the third column:

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

### Step 4: Using Row Operations
Next, we can perform row operations to simplify further. We will subtract the first column from the second column:

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

This gives us:
\[ D = -y \begin{vmatrix} x & a - x + y & 1 \\ y & b - y & 1 \\ z & c - z + y & 1 \end{vmatrix} \]

### Step 5: Evaluating the Determinant
Now we can expand the determinant. The determinant of a 3x3 matrix can be calculated using the formula:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:
\[ D = -y \left( x \cdot \begin{vmatrix} b - y & 1 \\ c - z + y & 1 \end{vmatrix} - (a - x + y) \cdot \begin{vmatrix} y & 1 \\ z & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} y & b - y \\ z & c - z + y \end{vmatrix} \right) \]

Calculating these determinants will yield the final result.

### Final Result
After performing the calculations and simplifying, we find that:
\[ D = -y (b - a)(c - b) \]

### Conclusion
Thus, the value of the determinant \( |(x, a+y, x + a), (y, b+y, y + b), (z, c+y, z+c)| \) is equal to:
\[ -y (b - a)(c - b) \]