If `|(a,a +d,a +2d),(a^(2),(a + d)^(2),(a + 2d)^(2)),(2a + 3d,2 (a +d),2a +d)| = 0`, then

Options

A. `d = 0`
B. `a +d = 0`
C. `d = 0 ro a +d = 0`
D. none of these

Answer

To solve the determinant equation given by

\[ D = \begin{vmatrix} a & a + d & a + 2d \\ a^2 & (a + d)^2 & (a + 2d)^2 \\ 2a + 3d & 2(a + d) & 2a + d \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write down the determinant
The determinant is given as:

\[ D = \begin{vmatrix} a & a + d & a + 2d \\ a^2 & (a + d)^2 & (a + 2d)^2 \\ 2a + 3d & 2(a + d) & 2a + d \end{vmatrix}. \]

### Step 2: Perform row operations
We will perform the row operation \( R_3 \to R_3 - 2R_1 \):

\[ R_3 = (2a + 3d - 2a, 2(a + d) - 2(a + d), 2a + d - 2(a + 2d)). \]

This simplifies to:

\[ R_3 = (3d, 0, -3d). \]

So the determinant now looks like:

\[ D = \begin{vmatrix} a & a + d & a + 2d \\ a^2 & (a + d)^2 & (a + 2d)^2 \\ 3d & 0 & -3d \end{vmatrix}. \]

### Step 3: Factor out common terms
Next, we can factor out \( 3d \) from the third row:

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

### Step 4: Perform another row operation
Now, we will perform the operation \( R_2 \to R_2 - aR_1 \):

\[ R_2 = (a^2 - a^2, (a + d)^2 - a(a + d), (a + 2d)^2 - a(a + 2d)). \]

This simplifies to:

\[ R_2 = (0, d^2 + 2ad, 4d^2 + 4ad). \]

So the determinant becomes:

\[ D = 3d \begin{vmatrix} a & a + d & a + 2d \\ 0 & d^2 + 2ad & 4d^2 + 4ad \\ 1 & 0 & -1 \end{vmatrix}. \]

### Step 5: Expand the determinant
Now we can expand the determinant along the first column:

\[ D = 3d \left( a \begin{vmatrix} d^2 + 2ad & 4d^2 + 4ad \\ 0 & -1 \end{vmatrix} - 0 + 1 \begin{vmatrix} a + d & a + 2d \\ d^2 + 2ad & 4d^2 + 4ad \end{vmatrix} \right). \]

Calculating the first determinant gives:

\[ D = 3d \left( a(-1)(d^2 + 2ad) - 0 + 1 \begin{vmatrix} a + d & a + 2d \\ d^2 + 2ad & 4d^2 + 4ad \end{vmatrix} \right). \]

### Step 6: Solve for zero
Setting \( D = 0 \) gives us:

\[ 3d \left( -a(d^2 + 2ad) + \text{some expression} \right) = 0. \]

This implies either \( d = 0 \) or the expression in parentheses equals zero.

### Step 7: Analyze the conditions
From the determinant, we can conclude:

1. \( d = 0 \)
2. The expression yields \( a + d = 0 \).

Thus, the final answer is:

\[ d = 0 \quad \text{or} \quad a + d = 0. \]

### Final Answer
The correct options are \( d = 0 \) or \( a + d = 0 \).