Answer
See below
Work Step by Step
We are given:
$A=\begin{bmatrix}
0 & x & y & z\\-x & 0 & 1 & -1\\-y & -1 & 0 & 1\\-z & 1 & -1 & 0
\end{bmatrix}\\
\rightarrow \det(A)=\begin{vmatrix}0 & x & y & z\\-x & 0 & 1 & -1\\-y & -1 & 0 & 1\\-z & 1 & -1 & 0\end{vmatrix}=0.\begin{vmatrix} 0 & 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{vmatrix}-x\begin{vmatrix}-x & 1 & -1\\-y & 0 & 1\\-z & -1 & 0\end{vmatrix}+y\begin{vmatrix}-x & 0 & -1\\-y & -1 & 1\\-z & 1 & 0\end{vmatrix}-z\begin{vmatrix}-x & 0 & 1\\-y & -1 & 0\\-z & 1 & -1 \end{vmatrix}$
The minors are:
$\begin{vmatrix}-x & 1 & -1\\-y & 0 & 1\\-z & -1 & 0\end{vmatrix}=-y-x-z$
$\begin{vmatrix}-x & 0 & -1\\-y & -1 & 1\\-z & 1 & 0\end{vmatrix}=x+y+z$
$\begin{vmatrix}-x & 0 & 1\\-y & -1 & 0\\-z & 1 & -1 \end{vmatrix}=-x-y-z$
Hence, $\det(A)=\begin{vmatrix}-x & 0 & 1\\-y & -1 & 0\\-z & 1 & -1 \end{vmatrix}\\=-x(-x-y-z)+y(x+y+z)-z(-s-y-z)\\=x(x+y+z)+y(x+y+z)+z(x+y+z)\\=(x+y+z)(x+y+z)\\=(x+y+z)^2$
