`|(a, a^2, 0), (1, 2a+b ,(a+b)), (0, 1, 2a+3b)|` is divisible by
a. `a+b`
b. `a+2b`
c. `2a+3b`
d. `a^2`

Options

A. `a+b `
B. `a+2b`
C. `2a+3b`
D. `a^(2)`

Explanation

To solve the determinant problem, we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:

\[ D = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a + b & a + b \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

### Step 2: Factor out 'a' from the first row
We can factor out \( a \) from the first row:

\[ D = a \cdot \begin{vmatrix} 1 & a & 0 \\ 1 & 2a + b & a + b \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

### Step 3: Perform Row Operations
Next, we perform the row operation \( R_2 \leftarrow R_2 - R_1 \):

\[ D = a \cdot \begin{vmatrix} 1 & a & 0 \\ 0 & (2a + b - a) & (a + b - 0) \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

This simplifies to:

\[ D = a \cdot \begin{vmatrix} 1 & a & 0 \\ 0 & (a + b) & (a + b) \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

### Step 4: Factor out \( (a + b) \) from the second row
Now, we can factor out \( (a + b) \) from the second row:

\[ D = a(a + b) \cdot \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

### Step 5: Perform another row operation
Next, we perform the row operation \( R_3 \leftarrow R_3 - R_2 \):

\[ D = a(a + b) \cdot \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & 1 \\ 0 & 0 & (2a + 3b - 1) \end{vmatrix} \]

### Step 6: Calculate the determinant
The determinant simplifies to:

\[ D = a(a + b)(2a + 3b - 1) \]

### Step 7: Identify divisibility
From the expression \( D = a(a + b)(2a + 3b - 1) \), we can see that \( D \) is divisible by \( a \) and \( a + b \).

### Conclusion
Thus, the determinant is divisible by:
- \( a \)
- \( a + b \)