If a, b, c, are in A.P, then the determinant `|(x+2,x+3,x+2a),( x+3,x+4,x+2b ),(x+4,x+5,x+2c)|` is
(A) `0` (B) `1` (C) `x` (D) `2x`

Explanation

To solve the determinant problem, we need to evaluate the determinant given the condition that \( a, b, c \) are in Arithmetic Progression (A.P).

### Step-by-Step Solution:

1. **Understanding the A.P Condition**:
Since \( a, b, c \) are in A.P, we can express \( b \) and \( c \) in terms of \( a \) and a common difference \( d \):
\[
b = a + d \quad \text{and} \quad c = a + 2d
...