= Solution
The cubic <Miller indices> give an <interplanar spacing> $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. Equivalently, the <reciprocal lattice vector> normal to these <lattice planes> has length $2\pi/d_{hkl}$. Hence
$$
\boxed{d_{111}=\frac{a}{\sqrt3},\qquad
\mathbf G_{111}=\frac{2\pi}{a}(1,1,1).}
$$
The eight $\{111\}$ <reciprocal lattice vectors> are the nearest nonzero reciprocal points. Their perpendicular bisectors form the eight hexagonal faces of the first <Brillouin zone>, a truncated octahedron. For the positive $(111)$ direction, that <Bragg plane> is $\mathbf k\cdot\mathbf G_{111}=|\mathbf G_{111}|^2/2$, or $k_x+k_y+k_z=3\pi/a$. It is parallel to the direct $(111)$ <lattice planes>. The six remaining square zone faces correspond to the $\{200\}$ reciprocal points. The zone boundary is halfway to a reciprocal point, not the point itself.
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