Let `Delta=|{:(a,a+b,a+3d),(a+d,a+2d,a),(a+2d, a, a+d):}|` then :

Options

A. `Delta` depends on a
B. `Delta` depends on d
C. `Delta` is independent of a,d
D. `Delta=0`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} a & a+b & a+3d \\ a+d & a+2d & a \\ a+2d & a & a+d \end{vmatrix} \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant as given:
\[ \Delta = \begin{vmatrix} a & a+b & a+3d \\ a+d & a+2d & a \\ a+2d & a & a+d \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can simplify the determinant by subtracting the first column from the second and third columns:
\[ \Delta = \begin{vmatrix} a & (a+b-a) & (a+3d-a) \\ a+d & (a+2d - (a+d)) & (a - (a+d)) \\ a+2d & (a - (a+2d)) & (a+d - (a+2d)) \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} a & b & 3d \\ a+d & d & -d \\ a+2d & -2d & -d \end{vmatrix} \]

### Step 3: Factor Out Common Terms
We can factor out \( d \) from the second and third columns:
\[ \Delta = d \begin{vmatrix} a & b & 3 \\ a+d & 1 & -1 \\ a+2d & -2 & -1 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we compute the determinant:
\[ \Delta = d \left( a \begin{vmatrix} 1 & -1 \\ -2 & -1 \end{vmatrix} - b \begin{vmatrix} a+d & -1 \\ a+2d & -1 \end{vmatrix} + 3 \begin{vmatrix} a+d & 1 \\ a+2d & -2 \end{vmatrix} \right) \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -1 \\ -2 & -1 \end{vmatrix} = (1)(-1) - (-1)(-2) = -1 - 2 = -3 \)
2. \( \begin{vmatrix} a+d & -1 \\ a+2d & -1 \end{vmatrix} = (a+d)(-1) - (-1)(a+2d) = -a - d + a + 2d = d \)
3. \( \begin{vmatrix} a+d & 1 \\ a+2d & -2 \end{vmatrix} = (a+d)(-2) - (1)(a+2d) = -2a - 2d - a - 2d = -3a - 4d \)

Putting it all together:
\[ \Delta = d \left( a(-3) - b(d) + 3(-3a - 4d) \right) \]
\[ = d \left( -3a - bd - 9a - 12d \right) \]
\[ = d \left( -12a - bd - 12d \right) \]

### Step 5: Final Expression
Thus, we have:
\[ \Delta = -d(12a + bd + 12d) \]

### Step 6: Conclusion
From the final expression, we can see that \( \Delta \) depends on \( b \) and \( d \), but not on \( a \). Therefore, the correct option is that \( \Delta \) depends on \( b \) and \( d \).