Since
$$
\left[\begin{array}{l}
1 \\
0 \\
1
\end{array}\right]+\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],
$$
we have
$$
A \mathbf{x}+\mathrm{Ay}=\mathrm{Az}
$$
It follows that we have
$$
\mathrm{A}(\mathbf{x}+\mathbf{y}-\mathbf{z})=\mathbf{0}
$$
Since the vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent, the linear combination $\mathbf{x}+\mathbf{y}-\mathbf{z} \neq \mathbf{0}$. Hence the matrix $\mathrm{A}$ is singular, and the determinant of $\mathrm{A}$ is zero.
(Recall that a matrix $\mathrm{A}$ is singular if and only if there exist nonzero vector $\mathbf{v}$ such that $A \mathbf{u}=\mathbf{0}$.)