I am not familiar with term mirror plane , hence I don't know how to solve this problem. As for operator itself, maybe if I select basis $(x,0,0), (0,y,0), (0,0,z)$ then I would express $x+z$ this way:
$ \mathcal{A}(x,0,0) = (1,0)$
$ \mathcal{A}(0,y,0) = (0,0)$
$ \mathcal{A}(0,0,z) = (0,1)$
Could you explain in detail how to solve this problem?
As I stated in the comment, I think the question is asking about the reflection through the plane given by $x + z = 0$.
That is, we take a vector, and reflect it through the plane perpendicular to the vector $(1,0,1)$.
Here is one way to solve the problem:
Note that the vectors $$ v_1 = (1,0,1), v_2 = (1,0,-1), v_3 = (0,1,0) $$ Form a basis of $\mathbb{R}^3$. In particular, we note that the plane $x + z = 0$ is given by the span of $\{v_2,v_3\}$, and the direction perpendicular to the plane is given by $v_1$.
Letting $A$ refer to the operator representing this reflection, we have $$ A(v_1) = -v_1\\ A(v_2) = v_2\\ A(v_3) = v_3 $$ So, letting $\mathcal B = \{v_1,v_2,v_3\}$, we note that $$ [A]_{\mathcal B \to \mathcal B} = \pmatrix{ -1&0&0\\0&1&0\\0&0&1 } $$