What is the value of the determinant ?
`|(x+1,x+2,x+4),(x+3,x+5,x+8),(x+7,x+10,x+14)|` ?

Options

A. x + 2
B. `x^(2) + 2`
C. 2
D. -2

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14 \end{vmatrix} \]

we will perform column operations to simplify the determinant.

### Step 1: Perform Column Operations
We will perform the following operations:
- \( C_2 \rightarrow C_2 - C_1 \)
- \( C_3 \rightarrow C_3 - C_1 \)

After applying these operations, the determinant becomes:

\[ D = \begin{vmatrix} x+1 & (x+2)-(x+1) & (x+4)-(x+1) \\ x+3 & (x+5)-(x+3) & (x+8)-(x+3) \\ x+7 & (x+10)-(x+7) & (x+14)-(x+7) \end{vmatrix} \]

Calculating the new values in the second and third columns:

- For \( C_2 \):
- \( x+2 - (x+1) = 1 \)
- \( x+5 - (x+3) = 2 \)
- \( x+10 - (x+7) = 3 \)

- For \( C_3 \):
- \( x+4 - (x+1) = 3 \)
- \( x+8 - (x+3) = 5 \)
- \( x+14 - (x+7) = 7 \)

Thus, the determinant simplifies to:

\[ D = \begin{vmatrix} x+1 & 1 & 3 \\ x+3 & 2 & 5 \\ x+7 & 3 & 7 \end{vmatrix} \]

### Step 2: Expand the Determinant
Now we will expand the determinant along the first row:

\[ D = (x+1) \begin{vmatrix} 2 & 5 \\ 3 & 7 \end{vmatrix} - 1 \begin{vmatrix} x+3 & 5 \\ x+7 & 7 \end{vmatrix} + 3 \begin{vmatrix} x+3 & 2 \\ x+7 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 5 \\ 3 & 7 \end{vmatrix} = (2)(7) - (5)(3) = 14 - 15 = -1 \)

2. \( \begin{vmatrix} x+3 & 5 \\ x+7 & 7 \end{vmatrix} = (x+3)(7) - (5)(x+7) = 7x + 21 - 5x - 35 = 2x - 14 \)

3. \( \begin{vmatrix} x+3 & 2 \\ x+7 & 3 \end{vmatrix} = (x+3)(3) - (2)(x+7) = 3x + 9 - 2x - 14 = x - 5 \)

Now substituting these back into the determinant expansion:

\[ D = (x+1)(-1) - 1(2x - 14) + 3(x - 5) \]

### Step 3: Simplify the Expression
Now we simplify:

\[ D = -(x + 1) - (2x - 14) + 3(x - 5) \]
\[ = -x - 1 - 2x + 14 + 3x - 15 \]
\[ = (-x - 2x + 3x) + (-1 + 14 - 15) \]
\[ = 0 + (-2) = -2 \]

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

\[ \boxed{-2} \]