Step-by-Step Solution

1. Apply the property of determinants: $C_1 \rightarrow C_1 + C_2$ $|\begin{matrix}x+y & y+z & z+x\\ z & x & y\\ 1 & 1 & 1\end{matrix}| = |\begin{matrix}x+y+y+z & y+z & z+x\\ z+x & x & y\\ 1+1 & 1 & 1\end{matrix}| = |\begin{matrix}x+2y+z & y+z & z+x\\ z+x & x & y\\ 2 & 1 & 1\end{matrix}|$
2. Apply the property of determinants: $R_1 \rightarrow R_1 - (x+y+z)R_3$ $|\begin{matrix}x+y & y+z & z+x\\ z & x & y\\ 1 & 1 & 1\end{matrix}| = |\begin{matrix}x+y-(x+y+z) & y+z-(x+y+z) & z+x-(x+y+z)\\ z & x & y\\ 1 & 1 & 1\end{matrix}| = |\begin{matrix}-z & -x & -y\\ z & x & y\\ 1 & 1 & 1\end{matrix}|$
3. Take -1 common from R1 $|\begin{matrix}-z & -x & -y\\ z & x & y\\ 1 & 1 & 1\end{matrix}| = -|\begin{matrix}z & x & y\\ z & x & y\\ 1 & 1 & 1\end{matrix}|$
4. Since R1 and R2 are identical, the value of the determinant is 0.