Without expanding,prove that $\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = 0$.

(A) Applying the row operation $R_{1} \rightarrow R_{1} + R_{2}$ to the determinant $\Delta$,we get: $\Delta = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$.
Taking $(x+y+z)$ as a common factor from $R_{1}$,we get: $\Delta = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$.
Since row $R_{1}$ and row $R_{3}$ are identical,the value of the determinant is $0$ according to the properties of determinants. Therefore,$\Delta = (x+y+z) \times 0 = 0$.