= Solution
Expand the two <linear combinations> and use the <Pauli matrix multiplication identity>:
$$
\begin{aligned}
\sigma(\mathbf a)\sigma(\mathbf b)
&=\sum_{ij}a_i b_j\left(\delta_{ij}I+i\sum_k\epsilon_{ijk}\sigma_k\right)\\
&=(\mathbf a\cdot\mathbf b)I+i\sigma(\mathbf a\times\mathbf b).
\end{aligned}
$$
Interchanging the <vectors> reverses the <cross product>, so their <commutator> is $2i\sigma(\mathbf a\times\mathbf b)$. The three <Pauli matrices> are <linearly independent>; alternatively, <trace> against $\sigma_k$ to recover each coefficient. Thus
$$
\boxed{[\sigma(\mathbf a),\sigma(\mathbf b)]=0\quad\Longleftrightarrow\quad\mathbf a\times\mathbf b=0.}
$$
For nonzero real <vectors>, this means they are parallel or antiparallel, equivalently <linearly dependent>. The algebra also extends to complex <vectors> using bilinear dot and cross products, with the same linear-dependence criterion; the <dot product> there has no complex conjugation.
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