The value of the determinant […] is:

The correct answer is Option (3) → 0

Apply R₁ → R₁ + R₂. Each entry of the new first row is \(x + y + z\):

    \( \displaystyle \Delta = \begin{vmatrix} x + y + z & x + y + z & x + y + z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \)

Take \(x + y + z\) common from R₁:

    \( \displaystyle \Delta = (x + y + z)\begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \)

Rows R₁ and R₃ are now identical, and a determinant with two identical rows is 0. So

    \(\Delta = 0\)