`Delta= |{:(1,,1+ac,,1+bc),(1,,1+ad,,1+bd),(1,,1+ae,,1+be):}|` is independent of

Options

A. a
B. b
C. c,d,e
D. none of these

Explanation

To solve the determinant given by

\[ \Delta = \begin{vmatrix} 1 & 1 + ac & 1 + bc \\ 1 & 1 + ad & 1 + bd \\ 1 & 1 + ae & 1 + be \end{vmatrix} \]

we will perform some operations to simplify the determinant.

### Step 1: Apply column operations
We will perform column operations to simplify the determinant. Specifically, we will subtract the first column from the second and third columns.

\[ C_2 \rightarrow C_2 - C_1 \quad \text{and} \quad C_3 \rightarrow C_3 - C_1 \]

This gives us:

\[ \Delta = \begin{vmatrix} 1 & (1 + ac) - 1 & (1 + bc) - 1 \\ 1 & (1 + ad) - 1 & (1 + bd) - 1 \\ 1 & (1 + ae) - 1 & (1 + be) - 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 & ac & bc \\ 1 & ad & bd \\ 1 & ae & be \end{vmatrix} \]

### Step 2: Factor out common terms
Now, we can factor out common terms from the second and third columns.

From column 2, we can factor out \(a\) and from column 3, we can factor out \(b\):

\[ \Delta = ab \begin{vmatrix} 1 & c & d \\ 1 & e & f \\ 1 & g & h \end{vmatrix} \]

### Step 3: Evaluate the determinant
Now, we notice that the first column is the same in all three rows. Therefore, the determinant will be zero because if two rows (or columns) of a determinant are identical, the value of the determinant is zero.

Thus, we have:

\[ \Delta = 0 \]

### Conclusion
Since the determinant \(\Delta\) is zero, it indicates that the variables \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) are independent. Therefore, the correct options for independence are:

- \(a\) and \(b\)
- \(c\), \(d\), and \(e\)