= Solution
The <parametric equation of a straight line> through $a$ with unit direction $\widehat t$ is
$$
\boxed{r=a+s\widehat t,\qquad s\in\mathbb R.}
$$
If the two lines meet, then $a_1+s\widehat t_1=a_2+t\widehat t_2$ for some real $s,t$, so $a_1-a_2$ lies in the <linear span> of their direction vectors. Its <dot product> with their <cross product> is consequently zero:
$$
\boxed{(a_1-a_2)\cdot(\widehat t_1\times\widehat t_2)=0.}
$$
This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and $\widehat t_1=\widehat t_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero <cross product>.
For nonparallel <skew lines>, put $n=(\widehat t_1\times\widehat t_2)/|\widehat t_1\times\widehat t_2|$. Every joining vector is $a_1-a_2+s\widehat t_1-t\widehat t_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^\perp$. This lower bound is attained, proving the <distance between skew lines>:
$$
\boxed{d=\frac{|(a_1-a_2)\cdot(\widehat t_1\times\widehat t_2)|}{|\widehat t_1\times\widehat t_2|}.}
$$
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