The value of the determinant `|(x+2,x+3,x+5),(x+4,x+6,x+9),(x+8,x+11,x+15)|` is

Options

A. 2
B. `-2`
C. 3
D. `x -1`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15 \end{vmatrix} \]

we can perform row operations to simplify the determinant.

### Step 1: Subtract Row 1 from Row 2 and Row 3
We will perform the following operations:
- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - R_2 \)

After performing these operations, we have:

\[ R_1 = (x+2, x+3, x+5) \]
\[ R_2 = ( (x+4)-(x+2), (x+6)-(x+3), (x+9)-(x+5) ) = (2, 3, 4) \]
\[ R_3 = ( (x+8)-(x+4), (x+11)-(x+6), (x+15)-(x+9) ) = (4, 5, 6) \]

Thus, the determinant now looks like:

\[ D = \begin{vmatrix} x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{vmatrix} \]

### Step 2: Subtract Row 2 from Row 1 and Row 3
Next, we will perform the following operations:
- \( R_1 \leftarrow R_1 - R_2 \)
- \( R_3 \leftarrow R_3 - R_2 \)

After performing these operations, we have:

\[ R_1 = ( (x+2)-2, (x+3)-3, (x+5)-4 ) = (x, x, x+1) \]
\[ R_3 = ( (4-2), (5-3), (6-4) ) = (2, 2, 2) \]

Thus, the determinant now looks like:

\[ D = \begin{vmatrix} x & x & x+1 \\ 2 & 3 & 4 \\ 2 & 2 & 2 \end{vmatrix} \]

### Step 3: Factor out common elements
We can factor out a 2 from Row 3:

\[ D = 2 \cdot \begin{vmatrix} x & x & x+1 \\ 2 & 3 & 4 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 4: Perform column operations
Now, we will perform the following operations:
- \( C_2 \leftarrow C_2 - C_1 \)
- \( C_3 \leftarrow C_3 - C_2 \)

After performing these operations, we have:

\[ D = 2 \cdot \begin{vmatrix} x & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 5: Expand the determinant
Now we can expand the determinant along the first column:

\[ D = 2 \cdot \left( x \cdot \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} \right) = 2 \cdot x \cdot (1 \cdot 0 - 1 \cdot 0) = 0 \]

### Final Result
Thus, the value of the determinant is:

\[ D = 0 \]