Let's first take it as read that the incident and reflected rays each travel in a straight line. You can prove this from Fermat's principle in one medium, Maxwell's equations in one medium or photonic $4$-momentum considerations, or just accept it as too obvious to need justifying.
When light strikes a point on a surface, it effectively strikes a tangential plane there. And in $\Bbb R^3$, given two lines and a plane not containing them that meet at a point, there's a second plane orthogonal to the first that contains both lines. We can work in this second plane hereafter.
So the real question is why angle of reflection equals angle of incidence. In particular, let's show why minimizing distance traversed (which is what Fermat's principle becomes in one medium) is equivalent to this angle equation. Suppose a ray incident from $A$ to a point $B$ on a reflecting surface subsequently traverses to $C$ (the points $A,\,C$ can be chosen arbitrarily on their respective rays, save for differing from $B$). Let $C$ have mirror image $C^\prime$ in the boundary, so $\vec{BC^\prime}$ is the mirror image of $\vec{BC}$. So$$\left|\vec{AB}\right|+\left|\vec{BC}\right|=\left|\vec{AB}\right|+\left|\vec{BC^\prime}\right|\ge\left|\vec{AC^\prime}\right|,$$with equality iff $B$ is where $\vec{AC^\prime}$ meets the boundary. You can do the angle-chasing to finish.
