Q.Show that the straight lines $x+1=2y=-12z$ and $x=y+2=6z-6$ are skew and hence find the shortest distance between them.

Two lines in space fall into exactly one of three relationships: intersecting (distance $0$), parallel (a fixed positive distance apart, never meeting), or skew — not parallel AND not meeting. Skew lines are precisely the pair that is not coplanar (they don't lie in any common plane), which is why the same $[\ ]$ machinery from the scalar-triple-product topic reappears here as a test.

Coplanarity / skewness test. For lines $\vec r=\vec a+s\vec b$ and $\vec r=\vec c+t\vec d$:

(If additionally $\vec b\parallel\vec d$, the lines are parallel rather than intersecting even though the test above gives $0$ — check parallelism first.)

Shortest distance, parallel lines (common direction $\vec b$): $d=\dfrac{|(\vec c-\vec a)\times\vec b|}{|\vec b|}$ — this is base $\times$ height turned sideways: $|\vec c-\vec a|\sin\theta$ where $\theta$ is the angle $\vec{AC}$ makes with the shared direction. …