If a, b, c are in A.P., then `|{:(x+2,x+3,x+a),(x+4, x+5, x+b),(x+6, x+7,x+c):}|` is

Answer

To solve the problem, we need to evaluate the determinant

\[ D = \begin{vmatrix} x + 2 & x + 3 & x + a \\ x + 4 & x + 5 & x + b \\ x + 6 & x + 7 & x + c \end{vmatrix} \]

given that \(a\), \(b\), and \(c\) are in Arithmetic Progression (A.P.).

### Step 1: Use the property of A.P.

Since \(a\), \(b\), and \(c\) are in A.P., we have the relationship:

\[ b - a = c - b \]

This can be rearranged to:

\[ 2b = a + c \quad \text{(1)} \]

### Step 2: Perform row operations

We will perform row operations to simplify the determinant. Specifically, we will subtract the second row from the first row and the third row from the second row.

1. **Row Operation 1**: \(R_1 \leftarrow R_1 - R_2\)

\[ R_1 = \begin{pmatrix} (x + 2) - (x + 4) & (x + 3) - (x + 5) & (x + a) - (x + b) \end{pmatrix} = \begin{pmatrix} -2 & -2 & a - b \end{pmatrix} \]

2. **Row Operation 2**: \(R_2 \leftarrow R_2 - R_3\)

\[ R_2 = \begin{pmatrix} (x + 4) - (x + 6) & (x + 5) - (x + 7) & (x + b) - (x + c) \end{pmatrix} = \begin{pmatrix} -2 & -2 & b - c \end{pmatrix} \]

Now the determinant becomes:

\[ D = \begin{vmatrix} -2 & -2 & a - b \\ -2 & -2 & b - c \\ x + 6 & x + 7 & x + c \end{vmatrix} \]

### Step 3: Analyze the rows

Notice that the first two rows are now:

- Row 1: \((-2, -2, a - b)\)
- Row 2: \((-2, -2, b - c)\)

Since \(b - a = c - b\) implies that \(a - b = b - c\), we can conclude that:

\[ a - b = b - c \quad \text{(from the A.P. condition)} \]

Thus, the first two rows are identical:

\[ R_1 = R_2 \]

### Step 4: Conclusion

According to the properties of determinants, if two rows are identical, the value of the determinant is zero.

Thus, we conclude:

\[ D = 0 \]

### Final Answer:

The value of the determinant is \(0\).

---