= Solution
{c}
Put $v=\mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3)\ne0$. Define
$$
\boxed{\mathbf b_1=\frac{2\pi}{v}\mathbf a_2\times\mathbf a_3,\quad\mathbf b_2=\frac{2\pi}{v}\mathbf a_3\times\mathbf a_1,\quad\mathbf b_3=\frac{2\pi}{v}\mathbf a_1\times\mathbf a_2.}
$$
The <scalar triple product> gives $\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}$. Consequently every integer combination $\mathbf G=\sum_jm_j\mathbf b_j$ satisfies
$$
\mathbf G\cdot\mathbf R=2\pi\sum_jm_jn_j\in2\pi\mathbb Z,\qquad e^{i\mathbf G\cdot\mathbf R}=1.
$$
This proves that the proposed vectors generate points in the <reciprocal lattice>, but primitiveness also requires the converse. Let $\mathbf G$ satisfy the exponential condition for every <lattice vector>, and in particular for $\mathbf R=\mathbf a_i$. Then
$$
m_i=\frac{\mathbf G\cdot\mathbf a_i}{2\pi}\in\mathbb Z.
$$
Since the $\mathbf b_i$ form a <basis> of three-dimensional space, the vector $\sum_i m_i\mathbf b_i$ has the same dot products with each $\mathbf a_j$ as $\mathbf G$. Their difference is orthogonal to three independent vectors and is zero. Hence
$$
\boxed{\Lambda^*=\left\{\sum_i m_i\mathbf b_i:m_i\in\mathbb Z\right\},}
$$
and no additional <reciprocal lattice vectors> lie outside that integer span. Thus the $\mathbf b_i$ are primitive, not merely three possible reciprocal translations.
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