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

Explanation

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 simplify the determinant using properties of determinants.

### Step 1: Apply Column Operations
We can simplify the determinant by performing column operations. Let's subtract the first column from the second and the third columns.

\[ C_2 \rightarrow C_2 - C_1 \]
\[ C_3 \rightarrow C_3 - C_1 \]

This gives us:

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

Calculating the new columns:

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

### Step 2: Apply Row Operations
Next, we will simplify the determinant further by performing row operations. Let's subtract the first row from the second and the third rows.

\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]

This gives us:

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

Calculating the new rows:

\[ D = \begin{vmatrix} x+1 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now we can calculate the determinant using the formula for a 3x3 matrix:

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

Where \( a = x+1, b = 1, c = 3 \), and the rest of the elements can be identified as follows:

\[ D = (x+1) \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 6 & 4 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 6 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = (1)(4) - (2)(2) = 4 - 4 = 0\)

2. \(\begin{vmatrix} 2 & 2 \\ 6 & 4 \end{vmatrix} = (2)(4) - (2)(6) = 8 - 12 = -4\)

3. \(\begin{vmatrix} 2 & 1 \\ 6 & 2 \end{vmatrix} = (2)(2) - (1)(6) = 4 - 6 = -2\)

Substituting back into the determinant formula:

\[ D = (x+1)(0) - 1(-4) + 3(-2) \]

This simplifies to:

\[ D = 0 + 4 - 6 = -2 \]

### Final Answer
Thus, the value of the determinant is

\[ \boxed{-2} \]