Q.Show that the lines $\dfrac{x-3}{3}=\dfrac{y-3}{-1},\ z-1=0$ and $\dfrac{x-6}{2}=\dfrac{z-1}{3},\ y-2=0$ intersect. Also find the point of intersection.

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. …