= Solution
The three <vectors> are <coplanar> precisely when their <scalar triple product> vanishes:
$$
\boxed{\mathbf a\cdot(\mathbf b\times\mathbf c)=0.}
$$
This includes <linearly dependent> or zero <vectors>. It expresses that the <determinant> with those <vectors> as columns is zero.
For the two moving particles, distinguish closest approach at a common time from the geometric <distance> between their supporting lines. For fixed reference times, put
$$
\mathbf d_0=\mathbf a_1-\mathbf a_2-\mathbf b_1t_1+\mathbf b_2t_2,\qquad
\mathbf v=\mathbf b_1-\mathbf b_2.
$$
Their relative position is $\mathbf d_0+t\mathbf v$. If $\mathbf v\ne0$, the <closest approach of two uniformly moving particles> occurs at
$$
\boxed{t_*=-\frac{\mathbf d_0\cdot\mathbf v}{|\mathbf v|^2},\qquad
d_{\min}^2=|\mathbf d_0|^2-\frac{(\mathbf d_0\cdot\mathbf v)^2}{|\mathbf v|^2}.}
$$
If $\mathbf v=0$, the <distance> is constant. If only $t\geq0$ is permitted, replace $t_*$ by $\max(0,t_*)$ before evaluating the <distance>.
Now suppose the velocities are nonparallel, and allow the relative starting-time offset to be chosen so that the particles reach the closest points of the two lines simultaneously. Write $\Delta=\mathbf a_2-\mathbf a_1$, $A=|\mathbf b_1|^2$, $B=\mathbf b_1\cdot\mathbf b_2$, $C=|\mathbf b_2|^2$ and $D=AC-B^2>0$. The independent <line> parameters $s_1,s_2$ at the closest points satisfy
$$
A s_1-Bs_2=\mathbf b_1\cdot\Delta,\qquad
B s_1-Cs_2=\mathbf b_2\cdot\Delta.
$$
These equations make the connecting <vector> perpendicular to both velocities. Solving the two-by-two <linear system> gives
$$
s_1=\frac{C\,\mathbf b_1\cdot\Delta-B\,\mathbf b_2\cdot\Delta}{D},\qquad
s_2=\frac{B\,\mathbf b_1\cdot\Delta-A\,\mathbf b_2\cdot\Delta}{D}.
$$
The corresponding <distance between skew lines> is $|\Delta\cdot(\mathbf b_1\times\mathbf b_2)|/|\mathbf b_1\times\mathbf b_2|$. Simultaneous arrival requires $t_1+s_1=t_2+s_2$, so $t_1-t_2=s_2-s_1$. Since the <vector triple product> gives
$$
(\mathbf b_1\times\mathbf b_2)\times(\mathbf b_1-\mathbf b_2)
=(A-B)\mathbf b_2+(C-B)\mathbf b_1,
$$
this condition is precisely
$$
\boxed{t_1-t_2=
\frac{\Delta\cdot[(\mathbf b_1\times\mathbf b_2)\times(\mathbf b_1-\mathbf b_2)]}
{B^2-AC}.}
$$
Thus the printed formula has a valid interpretation as a synchronization condition for attaining the line-distance lower bound. It is not an identity that must hold for arbitrary fixed reference times at the time of closest approach.
For a counterexample to that stronger reading, take $\mathbf a_1=0$, $\mathbf a_2=(1,0,1)$, $\mathbf b_1=(1,0,0)$, $\mathbf b_2=(0,1,0)$ and $t_1=t_2=0$. The actual squared <distance> is $2(t-1/2)^2+3/2$, minimized at $t=1/2$. The printed right-hand side is $-1$, whereas the fixed difference $t_1-t_2$ is zero. The line-distance bound is one, but it is not attained by that synchronized motion. Parallel velocities also require separate treatment because the displayed denominator then vanishes.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-resit-closest-approach.png]
{title=Closest line points need not be reached at the same time; shifting a trajectory's time origin changes the actual minimum distance}
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