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

Using the shortest-distance-between-parallel-lines formula, the perpendicular distance works out exactly to $3$.

1. Line 1: $\dfrac{x-3}{4}=\dfrac{y-1}{2}=\dfrac{z-5}{-3}$ passes through $A_1(3,1,5)$ with direction $\vec b=4\vec i+2\vec j-3\vec k$.
2. Line 2: $\dfrac{x-1}{4}=\dfrac{y-2}{2}=\dfrac{z-3}{-3}$ passes through $A_2(1,2,3)$ with the same direction $\vec b$, confirming the lines are parallel.
3. Vector joining the two points: $\vec{A_1A_2}=(1-3)\vec i+(2-1)\vec j+(3-5)\vec k=-2\vec i+\vec j-2\vec k$.
4. Shortest distance between parallel lines: $d=\dfrac{\left|\vec{A_1A_2}\times\vec b\right|}{|\vec b|}$.
5. Compute the cross product: $\vec{A_1A_2}\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\-2&1&-2\\4&2&-3\end{vmatrix}=(1)(1\vec i)+\ldots$ Component-wise: $i$-comp $=(1)(-3)-(-2)(2)=-3+4=1$; $j$-comp $=-[(-2)(-3)-(-2)(4)]=-[6+8]=-14$; $k$-comp $=(-2)(2)-(1)(4)=-4-4=-8$. So the cross product is $(1,-14,-8)$. …