= Solution
Using <Einstein notation> and
$$
\epsilon_{ijk}\epsilon_{klm}
=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl},
$$
we obtain
$$
\begin{aligned}
[a\times(b\times c)]_i
&=\epsilon_{ijk}a_j\epsilon_{klm}b_lc_m\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})
a_jb_lc_m\\
&=(a\cdot c)b_i-(a\cdot b)c_i.
\end{aligned}
$$
Thus
$$
\boxed{a\times(b\times c)=(a\cdot c)b-(a\cdot b)c}.
$$
The <scalar triple product> is
$$
[a,b,c]=a\cdot(b\times c).
$$
Put $\Delta=[a,b,c]$. When $\Delta\ne0$, the three vectors
$$
e'_1=\frac{b\times c}{\Delta},
\qquad
e'_2=\frac{c\times a}{\Delta},
\qquad
e'_3=\frac{a\times b}{\Delta}
$$
form the <reciprocal basis> to $a,b,c$, so $e'_i\cdot e_j=\delta_{ij}$. Their scalar triple product is $1/\Delta$. Therefore
$$
[a\times b,b\times c,c\times a]
=\Delta^3[e'_3,e'_1,e'_2]
=\boxed{\Delta^2},
$$
since the cyclic permutation preserves orientation. The identity also holds when $\Delta=0$, either by continuity or directly because the cross products are then linearly dependent.
For the given basis $e_1,e_2,e_3$, the same identities give
$$
e'_i\cdot e_j=\delta_{ij}.
$$
If $\sum_i\alpha_ie'_i=0$, dotting with $e_j$ gives $\alpha_j=0$, so the $e'_i$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e'_j=\delta_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives
$$
\boxed{e''_i=e_i\qquad(i=1,2,3)}.
$$
Every vector $K$ has a unique expansion $K=\sum_i\kappa_ie'_i$, and then
$$
K\cdot R
=\sum_{i,j}\kappa_i n_j e'_i\cdot e_j
=\sum_i\kappa_i n_i.
$$
This is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every $\kappa_i$ is an integer. Hence all such points are
$$
\boxed{K=m_1e'_1+m_2e'_2+m_3e'_3,
\qquad m_1,m_2,m_3\in\mathbb Z}.
$$
They form the reciprocal lattice in the convention without a factor of $2\pi$.
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