There are a few ways to approach this, but they all hinge on two key properties of reflection. First is Snell’s law of reflection: the angle of incidence is equal to the angle of reflection. The second is that a ray through $\mathbf A$ and its reflection from $\mathbf x$ lie in a plane through $\mathbf A$ that is perpendicular to $\mathbf x$. Using these properties, one can attack the problem in a few slightly different ways.
There’s also a trick that you can use to simplify many of these reflection problems: if you extend the incident ray whose reflection passes through $\mathbf A$ beyond the mirror surface, it will pass through the reflection (used in a different sense) of $\mathbf A$ on the other side of the mirror plane $\mathbf x$. This trick is an application of the above properties.
So, for this particular problem, I would proceed by finding the reflection $\mathbf A'$ of $\mathbf A$ in the plane $\mathbf x$, and then finding a perpendicular plane through $\mathbf A'$ such that its intersections with the two lines and $\mathbf A'$ are colinear. From there, computing the two rays and the point of reflection is a straightforward computation.