If `a-2b+c=1`, then the value of `|{:(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c):}|` is

Options

A. (a)`x`
B. (b)`-x`
C. (c)`-1`
D. (d)`1`

Explanation

To solve the given determinant, we start with the expression:

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

### Step 1: Apply Row Operations
We will perform row operations to simplify the determinant. Let's modify Row 1 by applying the operation \( R_1 \rightarrow R_1 - 2R_2 + R_3 \).

\[ R_1 = \begin{pmatrix} x + 1 - 2(x + 2) + (x + 3) & x + 2 - 2(x + 3) + (x + 4) & x + a - 2(x + b) + (x + c) \end{pmatrix} \]

Calculating each element:

- First element:
\[ x + 1 - 2(x + 2) + (x + 3) = x + 1 - 2x - 4 + x + 3 = 0 \]

- Second element:
\[ x + 2 - 2(x + 3) + (x + 4) = x + 2 - 2x - 6 + x + 4 = 0 \]

- Third element:
\[ x + a - 2(x + b) + (x + c) = x + a - 2x - 2b + x + c = a - 2b + c \]

Thus, Row 1 becomes:
\[ R_1 = \begin{pmatrix} 0 & 0 & a - 2b + c \end{pmatrix} \]

So the determinant now looks like:
\[ D = \begin{vmatrix} 0 & 0 & a - 2b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} \]

### Step 2: Substitute the Given Condition
From the problem, we know that \( a - 2b + c = 1 \). Thus, we can substitute this into our determinant:

\[ D = \begin{vmatrix} 0 & 0 & 1 \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} \]

### Step 3: Expand the Determinant
We can expand the determinant along the first row. Since the first two elements of the first row are zero, we only need to consider the last element:

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

Calculating this 2x2 determinant:
\[ \begin{vmatrix} x + 2 & x + 3 \\ x + 3 & x + 4 \end{vmatrix} = (x + 2)(x + 4) - (x + 3)(x + 3) \]

Expanding:
\[ = (x^2 + 6x + 8) - (x^2 + 6x + 9) = 8 - 9 = -1 \]

### Final Result
Thus, the value of the determinant \( D \) is:
\[ D = 1 \cdot (-1) = -1 \]

### Conclusion
The value of the determinant is:
\[ \boxed{-1} \]