= Solution
{c}
Given independent <primitive lattice translation vectors> $\mathbf a_1,\mathbf a_2,\mathbf a_3$, define
$$
V_p=\mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3),\qquad
\mathbf b_1=2\pi\frac{\mathbf a_2\times\mathbf a_3}{V_p},\quad
\mathbf b_2=2\pi\frac{\mathbf a_3\times\mathbf a_1}{V_p},\quad
\mathbf b_3=2\pi\frac{\mathbf a_1\times\mathbf a_2}{V_p}.
$$
Then $\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}$. The <reciprocal lattice> consists of
$$
\boxed{\mathbf G=h\mathbf b_1+k\mathbf b_2+l\mathbf b_3,\qquad h,k,l\in\mathbb Z,}
$$
equivalently all <wavevectors> for which $e^{i\mathbf G\cdot\mathbf R}=1$ for every direct <lattice vector> $\mathbf R$.
For the <face-centered cubic lattice>, choose
$$
\mathbf a_1=\frac a2(0,1,1),\quad
\mathbf a_2=\frac a2(1,0,1),\quad
\mathbf a_3=\frac a2(1,1,0).
$$
Their <scalar triple product> is $a^3/4$, and direct evaluation of the <cross products> gives
$$
\mathbf b_1=\frac{2\pi}{a}(-1,1,1),\quad
\mathbf b_2=\frac{2\pi}{a}(1,-1,1),\quad
\mathbf b_3=\frac{2\pi}{a}(1,1,-1).
$$
Every integer combination has coordinates $(2\pi/a)(H,K,L)$ with $H,K,L$ all even or all odd. Conversely, for such coordinates the coefficients $(K+L)/2,(H+L)/2,(H+K)/2$ are integers, so every such point occurs. The even-coordinate points are the corners of cubes of edge $4\pi/a$, and the odd-coordinate points are their body centres. Therefore
$$
\boxed{\text{reciprocal Bravais lattice: body-centred cubic},\qquad a_{\rm reciprocal}=\frac{4\pi}{a}.}
$$
This is a property of the <Bravais lattice>; the two-atom <atomic basis> changes <structure factors> rather than these <reciprocal lattice> translation vectors.
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