If `Delta=|(x_1+iy_1,x_1i-y_1,z_1),(x_2+iy_2,x_2i-y_2,z_2),(x_3+iy_3,x_3i-y_3,z_3)|` then `Delta` is

Options

A. `-1`
B. 2i
C. 0
D. 3i

Answer

To solve the determinant given by

\[ \Delta = \begin{vmatrix} x_1 + iy_1 & x_1 i - y_1 & z_1 \\ x_2 + iy_2 & x_2 i - y_2 & z_2 \\ x_3 + iy_3 & x_3 i - y_3 & z_3 \end{vmatrix} \]

we can apply properties of determinants to simplify the calculation.

### Step 1: Break down the first column
We can separate the first column into real and imaginary parts:

\[ \Delta = \begin{vmatrix} x_1 + iy_1 & x_1 i - y_1 & z_1 \\ x_2 + iy_2 & x_2 i - y_2 & z_2 \\ x_3 + iy_3 & x_3 i - y_3 & z_3 \end{vmatrix} \]

This can be rewritten as:

\[ \Delta = \begin{vmatrix} x_1 & x_1 i - y_1 & z_1 \\ x_2 & x_2 i - y_2 & z_2 \\ x_3 & x_3 i - y_3 & z_3 \end{vmatrix} + i \begin{vmatrix} y_1 & x_1 i - y_1 & z_1 \\ y_2 & x_2 i - y_2 & z_2 \\ y_3 & x_3 i - y_3 & z_3 \end{vmatrix} \]

### Step 2: Factor out common terms
For the first determinant, we can factor out \(i\) from the second column:

\[ \Delta = \begin{vmatrix} x_1 & x_1 i & z_1 \\ x_2 & x_2 i & z_2 \\ x_3 & x_3 i & z_3 \end{vmatrix} - \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 3: Evaluate the determinants
The first determinant can be simplified by taking \(i\) out:

\[ \Delta = i \begin{vmatrix} x_1 & x_2 & x_3 \\ x_1 & x_2 & x_3 \\ z_1 & z_2 & z_3 \end{vmatrix} - \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 4: Recognize properties of determinants
Notice that the first determinant has two identical columns, which means its value is zero:

\[ \begin{vmatrix} x_1 & x_2 & x_3 \\ x_1 & x_2 & x_3 \\ z_1 & z_2 & z_3 \end{vmatrix} = 0 \]

Thus, we have:

\[ \Delta = 0 - \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 5: Final result
The final result for \(\Delta\) is:

\[ \Delta = -\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Conclusion
Thus, the value of \(\Delta\) is determined by the negative of the determinant of the matrix formed by the real parts and imaginary parts of the original columns.