Let `Delta=|{:(a,a+d,a+3d),(a+d,a+2d,a),(a+2d, a, a+d):}|` then :
i) Δ depends on a
ii)Δ depends on d
iii)Δ is independent of a,d
iv)Δ = 0

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

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

### Step 2: Simplify the determinant
We can perform row operations to simplify the determinant. Let's subtract the first row from the second and the third rows:
\[ \Delta = \begin{vmatrix} a & a+d & a+3d \\ (a+d - a) & (a+2d - (a+d)) & (a - (a+3d)) \\ (a+2d - a) & (a - (a+d)) & (a+d - (a+3d)) \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} a & a+d & a+3d \\ d & d & -3d \\ 2d & -d & -2d \end{vmatrix} \]

### Step 3: Factor out common terms
Next, we can factor out \(d\) from the second and third rows:
\[ \Delta = d \cdot d \cdot \begin{vmatrix} a & a+d & a+3d \\ 1 & 1 & -3 \\ 2 & -1 & -2 \end{vmatrix} \]
This gives us:
\[ \Delta = d^2 \begin{vmatrix} a & a+d & a+3d \\ 1 & 1 & -3 \\ 2 & -1 & -2 \end{vmatrix} \]

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

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 1 & -3 \\ -1 & -2 \end{vmatrix} = (1)(-2) - (-3)(-1) = -2 - 3 = -5\)
2. \(\begin{vmatrix} 1 & -3 \\ 2 & -2 \end{vmatrix} = (1)(-2) - (-3)(2) = -2 + 6 = 4\)
3. \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3\)

Putting it all together:
\[ \Delta = d^2 \left( a(-5) - (a+d)(4) + (a+3d)(-3) \right) \]
Expanding this:
\[ \Delta = d^2 \left( -5a - 4a - 4d - 3a - 9d \right) = d^2 \left( -12a - 13d \right) \]

### Step 5: Conclusion
Thus, we have:
\[ \Delta = -d^2(12a + 13d) \]

### Final Result
From this expression, we can conclude:
- \(\Delta\) depends on \(d\) and \(a\).
- \(\Delta\) is not independent of \(a\) and \(d\).
- \(\Delta\) is not equal to zero unless both \(d\) and \(a\) are zero.