= Solution
In an oriented orthonormal coordinate system, the <dot product> and <cross product> are
$$
a\cdot b=\sum_{k=1}^3a_kb_k,\qquad
a\times b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1).
$$
The <scalar triple product> is $a\cdot(b\times c)$; equivalently it is the <determinant> of the <matrix> whose columns are $a,b,c$. The three vectors have <linear independence> when
$$
t_1a_1+t_2a_2+t_3a_3=0\ \Longrightarrow\ t_1=t_2=t_3=0.
$$
Let $U=(a_1\ a_2\ a_3)$ and $V=(b_1\ b_2\ b_3)$ be column <matrices>. Since $(U^TV)_{ij}=a_i\cdot b_j$, we have $S=U^TV$. Multiplicativity of the <determinant> and its invariance under <matrix transpose> yield
$$
\det S=\det(U^T)\det V=\det U\det V
=\boxed{\bigl(a_1\cdot(a_2\times a_3)\bigr)
\bigl(b_1\cdot(b_2\times b_3)\bigr)}.
$$
A square <matrix> has maximal <matrix rank> exactly when its <determinant> is nonzero, equivalently when its columns have <linear independence>. A product of two real numbers is nonzero exactly when both factors are nonzero. Hence $\boxed{\operatorname{rank}S=3\iff (a_i)_{i=1}^3\text{ and }(b_j)_{j=1}^3\text{ are both independent}}$.
The same argument works in every dimension. Set $U_n=(c_1\ \cdots\ c_n)$ and $V_n=(d_1\ \cdots\ d_n)$. Then
$$
T=U_n^TV_n,\qquad \det T=\det U_n\det V_n,
$$
so $\boxed{\operatorname{rank}T=n\iff(c_i)_{i=1}^n\text{ and }(d_j)_{j=1}^n\text{ are both independent}}$.
For the last construction, choose two <standard basis> vectors $e_1,e_2$ in $\mathbb R^n$ and take $c_j=e_1+je_2$ for $j=1,\ldots,n$. If $\alpha c_i+\beta c_j=0$ with $i\ne j$, its first two coordinates give $\alpha+\beta=0$ and $i\alpha+j\beta=0$, hence $\alpha=\beta=0$. Every pair therefore has <linear independence>. Every triple lies in the two-dimensional <linear span> of $e_1,e_2$ and has <linear dependence>. More explicitly, for distinct $i,j,k$,
$$
(j-k)c_i+(k-i)c_j+(i-j)c_k=0,
$$
a nontrivial dependence. Thus $\boxed{c_j=e_1+je_2\ (1\leq j\leq n)\text{ works for every }n>2}$.
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