Past exam of the materials science course of the University of Cambridge / 2016 / Paper 1 / 1 / a / iii / Solution

The relevant planes contain $[100]$, so their normals lie in the $bc$ plane. The symmetry operations can be checked directly on the height-labelled molecules:

Reflection reverses the bond slope, and the $c/2$ translation carries A to B with the correct height.

This again exchanges A and B. Combining it with the C-centering translation gives the equivalent description with a $b/2$ glide in the same plane.

There are no ordinary mirror planes containing $[100]$ for the illustrated generic bond inclination. Reflection alone would reverse a bond slope without the necessary accompanying translation. A mirror perpendicular to $[100]$ is a different symmetry and is outside the requested zone.