 Solution

Put $v={\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)\ne 0$. DefineThe scalar triple product gives ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. Consequently every integer combination $\mathbf{G}={\sum }_j{m}_j{\mathbf{b}}_j$ satisfiesThis 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 $\mathbf{R}={\mathbf{a}}_i$. ThenSince 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. Henceand no additional reciprocal lattice vectors lie outside that integer span. Thus the ${\mathbf{b}}_i$ are primitive, not merely three possible reciprocal translations.