Suppose `a,b,c,x,y epsilon R`. Let
`Delta=|(1,2+ax,3+ay),(1,2+bx,3+by),(1,2+cx,3+cy)|`
Then `Delta` is independnet is

Options

A. a,b,c
B. x,y
C. a,b,c,y
D. a,b,c,x,y

Answer

To solve the problem, we need to analyze the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} 1 & 2 + ax & 3 + ay \\ 1 & 2 + bx & 3 + by \\ 1 & 2 + cx & 3 + cy \end{vmatrix} \]

### Step 1: Simplifying the Determinant
We can simplify the determinant by performing column operations. We will replace the second column with the second column minus \( 2 \) times the first column and the third column with the third column minus \( 3 \) times the first column.

\[ \Delta = \begin{vmatrix} 1 & (2 + ax) - 2 \cdot 1 & (3 + ay) - 3 \cdot 1 \\ 1 & (2 + bx) - 2 \cdot 1 & (3 + by) - 3 \cdot 1 \\ 1 & (2 + cx) - 2 \cdot 1 & (3 + cy) - 3 \cdot 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 & ax & ay \\ 1 & bx & by \\ 1 & cx & cy \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Next, we can factor out \( x \) from the second column and \( y \) from the third column:

\[ \Delta = xy \begin{vmatrix} 1 & a & b \\ 1 & b & b \\ 1 & c & c \end{vmatrix} \]

### Step 3: Identifying Proportional Columns
Now we can observe that the first column is identical in all rows, which leads us to conclude that the second and third columns are proportional:

\[ \Delta = xy \begin{vmatrix} 1 & a & b \\ 1 & b & b \\ 1 & c & c \end{vmatrix} \]

Since the first column is the same for all rows, the determinant will be zero:

\[ \Delta = 0 \]

### Step 4: Conclusion
Since \( \Delta = 0 \), it is independent of \( a, b, c, x, \) and \( y \). The determinant being zero indicates that the rows (or columns) are linearly dependent.

Thus, the final answer is that \( \Delta \) is independent of \( a, b, c, x, y \).