Let `Delta(x)=|(x+a,x+b,x+a-c),(x+b,x+c,x-1),(x+c,x+d,x-b+d)| and int_0^2 Delta(x)dx=-16,` where `a,b,c,d` are in A.P. then the common difference (i) `1` (ii)`2` (iii)`3` (iv)`4`

Options

A. `+-1`
B. `+-2`
C. `+-3`
D. `+-4`

Answer

To solve the problem, we need to evaluate the determinant \( \Delta(x) \) and then use the given integral condition to find the common difference \( d \) of the arithmetic progression (A.P.) formed by \( a, b, c, d \).

### Step 1: Write the determinant \( \Delta(x) \)
The determinant is given by:
\[ \Delta(x) = \begin{vmatrix} x + a & x + b & x + a - c \\ x + b & x + c & x - 1 \\ x + c & x + d & x - b + d \end{vmatrix} \]

### Step 2: Apply row operations to simplify the determinant
We will perform row operations to simplify the determinant.

1. **Row 2**: Replace it with \( \text{Row 2} - \text{Row 1} \).
2. **Row 3**: Replace it with \( \text{Row 3} - \text{Row 2} \).

After performing these operations, we have:
\[ \Delta(x) = \begin{vmatrix} x + a & x + b & x + a - c \\ b - a & c - b & -1 - (a - c) \\ (c - b) & (d - c) & (-b + d + 1) \end{vmatrix} \]

### Step 3: Further simplify the determinant
Let’s denote the differences:
- Let \( d = b - a \)
- Let \( c - b = d \)
- Let \( d - c = d \)

Now we can rewrite the determinant as:
\[ \Delta(x) = \begin{vmatrix} x + a & x + b & x + a - c \\ d & d & -1 - (a - c) \\ d & d & (-b + d + 1) \end{vmatrix} \]

### Step 4: Calculate the determinant
Using properties of determinants:
\[ \Delta(x) = (x + a) \begin{vmatrix} d & -1 - (a - c) \\ d & (-b + d + 1) \end{vmatrix} - (x + b) \begin{vmatrix} d & -1 - (a - c) \\ d & (-b + d + 1) \end{vmatrix} + (x + a - c) \begin{vmatrix} d & d \\ d & d \end{vmatrix} \]
The last determinant is zero since the rows are identical.

### Step 5: Evaluate the integral
Given that:
\[ \int_0^2 \Delta(x) \, dx = -16 \]
From the determinant, we found that \( \Delta(x) = -2d^2 \). Therefore:
\[ \int_0^2 -2d^2 \, dx = -16 \]
Calculating the integral:
\[ -2d^2 \cdot (2 - 0) = -4d^2 = -16 \]
Thus, we have:
\[ 4d^2 = 16 \implies d^2 = 4 \implies d = \pm 2 \]

### Conclusion
The common difference \( d \) can be either \( 2 \) or \( -2 \). Since we are asked for the positive common difference, the answer is:
\[ \boxed{2} \]