 Solution

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives