Let `x_1y_1z_1,x_2y_2z_2 and x_3y_3z_3` be three 3-digit even numbers and `Delta=|{:(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3):}|`. then `Delta` is

Options

A. divisible by 2 but not necessarily by 4
B. divisible by 4 but not necessarily by 8
C. divisible by 8
D. none of these

Answer

To find the value of \(\Delta = |(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)|\), we will compute the determinant of the matrix formed by the coordinates of the three 3-digit even numbers.

### Step 1: Write the matrix
The determinant \(\Delta\) can be represented as:
\[ \Delta = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 2: Apply the determinant formula
The formula for the determinant of a 3x3 matrix is given by:
\[ \Delta = x_1(y_2z_3 - y_3z_2) - y_1(x_2z_3 - x_3z_2) + z_1(x_2y_3 - x_3y_2) \]

### Step 3: Substitute the values
Since \(x_1, y_1, z_1, x_2, y_2, z_2, x_3, y_3, z_3\) are all 3-digit even numbers, we can substitute these values into the determinant formula. However, since they are even numbers, we can conclude that:
- Each \(x_i\), \(y_i\), and \(z_i\) can be expressed in the form \(2k\) where \(k\) is an integer.

### Step 4: Factor out the common factor
Since all the numbers are even, we can factor out \(2\) from each term:
\[ \Delta = 2 \cdot \begin{vmatrix} k_1 & m_1 & n_1 \\ k_2 & m_2 & n_2 \\ k_3 & m_3 & n_3 \end{vmatrix} \]
where \(k_i = \frac{x_i}{2}, m_i = \frac{y_i}{2}, n_i = \frac{z_i}{2}\).

### Step 5: Conclusion
The determinant \(\Delta\) is thus an even number because it is a multiple of \(2\). The exact value of \(\Delta\) will depend on the specific values of \(x_i, y_i, z_i\), but it will always be even.

### Final Result
Therefore, \(\Delta\) is an even number.

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