= Solution
The sum of the three primitive vectors is $(a/2)(1,1,1)$, so the surface is a $(111)$ plane. Two in-plane translations are
$$
t_1=a(1,-1,0),\qquad t_2=a(1,0,-1).
$$
They have length $L=\sqrt2a$ and included angle $60^\circ$, giving a triangular top-layer lattice. Body-centred sites lie in offset parallel layers, not as additional sites in this same plane. A primitive surface cell is the rhombus spanned by $t_1,t_2$, of area
$$
\boxed{A_s=|t_1\times t_2|=\sqrt3a^2=17.3503\ \text{\AA}^2}.
$$
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-2-tungsten-surface.png]
{title=Triangular top atomic layer of bcc tungsten (111) and its hexagonal surface Brillouin zone}
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