$\Delta=\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|$

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y\end{array}\right|$

$=2(x+y+z)\left|\begin{array}{ccc}1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:

$\Delta=2(x+y+z)\left|\begin{array}{ccc}1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z\end{array}\right|$

$=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

Expanding along $R_{3},$ we have:

$\Delta=2(x+y+z)^{3}(1)(1-0)=2(x+y+z)^{3}$

Hence, the given result is proved.