if `D=|{:(x+d,x+e,x+f),(x+d+1,x+e+1,x+f+1),(x+a,x+b,x+c):}|` then D does not depend on :

Options

A. a
B. e
C. d
D. x

Answer

To solve the problem, we need to evaluate the determinant \( D \) given by the matrix:

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

### Step 1: Write down the determinant
We start with the determinant as given:

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

### Step 2: Perform row operation
We will subtract the first row from the second row:

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

### Step 3: Perform column operations
Next, we will perform column operations. We subtract the second column from the first column and the third column from the second column:

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

This simplifies to:

\[ D = \begin{vmatrix} d - e & e - f & f \\ 0 & 0 & 1 \\ a - b & b - c & c \end{vmatrix} \]

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

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

### Step 5: Calculate the determinant
The determinant can be calculated as follows:

\[ D = -1 \cdot \left( (d - e)(b - c) - (e - f)(a - b) \right) \]

### Conclusion
Notice that the expression does not contain \( x \) at all. Therefore, we conclude that \( D \) does not depend on \( x \).

Thus, the final answer is that \( D \) does not depend on \( x \).