If x,y, z are in A.P., then the values of the determinant `|(a +2,a +3,a + 2y),(a + 3,a +4,a + 2y),(a +4,a +5,a + 2z)|`, is

Options

A. 1
B. 0
C. 2a
D. a

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} a + 2 & a + 3 & a + 2y \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]

given that \(x, y, z\) are in arithmetic progression (A.P.), we proceed with the following steps:

### Step 1: Understand the condition of A.P.
Since \(x, y, z\) are in A.P., we have the relation:
\[ 2y = x + z \]

### Step 2: Write the determinant
We can express the determinant as:
\[ D = \begin{vmatrix} a + 2 & a + 3 & a + 2y \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]

### Step 3: Apply row transformation
We will perform a row operation to simplify the determinant. Let's transform \(R_1\) by adding \(R_3\) to it:
\[ R_1 \rightarrow R_1 + R_3 \]
This gives us:
\[ D = \begin{vmatrix} (2a + 6) & (2a + 8) & (a + 2y + a + 2z) \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]

### Step 4: Simplify the determinant
Now, we can express \(a + 2y + a + 2z\) as:
\[ a + 2y + a + 2z = 2a + 2y + 2z = 2a + 2(x + z) = 2a + 2(2y) = 2a + 4y \]
Thus, the determinant becomes:
\[ D = \begin{vmatrix} 2a + 6 & 2a + 8 & 2a + 4y \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]

### Step 5: Factor out common terms
Next, we can factor out a 2 from the first row:
\[ D = 2 \begin{vmatrix} a + 3 & a + 4 & a + 2y \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]

### Step 6: Identify duplicate rows
Notice that the first two rows of the determinant are now identical:
\[ \begin{vmatrix} a + 3 & a + 4 & a + 2y \\ a + 3 & a + 4 & a + 2y \\ a + 4 & a + 5 & a + 2z \end{vmatrix} \]
When two rows of a determinant are the same, the value of the determinant is zero.

### Final Answer
Thus, the value of the determinant \(D\) is:
\[ D = 0 \]