Q.Show that the lines $\vec r=(6\hat i+\hat j+2\hat k)+s(\hat i+2\hat j-3\hat k)$ and $\vec r=(3\hat i+2\hat j-2\hat k)+t(2\hat i+4\hat j-5\hat k)$ are skew lines 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.

Shortest distance, skew lines: $\delta=\left|\dfrac{(\vec c-\vec a)\cdot(\vec b\times\vec d)}{|\vec b\times\vec d|}\right|$ — exactly the numerator of the coplanarity test, divided by $|\vec b\times\vec d|$ to normalise it into an actual length. Geometrically, $\vec b\times\vec d$ is the unique direction perpendicular to BOTH lines, and $\delta$ is the length of $\vec{AC}$'s projection onto that direction.

Foot of a perpendicular from a point $D$ to a line $\vec r=\vec a+t\vec b$: write the foot as the line's general point $F=\vec a+t\vec b$, form $\vec{DF}$, and solve $\vec b\cdot\vec{DF}=0$ (perpendicularity) for $t$; substitute back for $F$, and $|\vec{DF}|$ is the perpendicular distance.