HMMT-Nov 2015 guts, Problem 28

Find the shortest distance between the lines $\frac{x+2}{2}=\frac{y-1}{3}=\frac{z}{1}$ and $\frac{x-3}{-1}=\frac{y}{1}=\frac{z+1}{2}$

Solution: First we find the direction of a line perpendicular to both of these lines. By taking the cross product $(2,3,1) \times(-1,1,2)=(5,-5,5)$ we find that the plane $x-y+z+3=0$ contains the first line and is parallel to the second. Now we take a point on the second line, say the point $(3,0,-1)$ and find the distance between this point and the plane. This comes out to $\frac{|3-0+(-1)+3|}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{5}{\sqrt{3}}=\frac{5 \sqrt{3}}{3}$. $\fbox{\frac{5 \sqrt{3}}{3}}$.

Answer: \frac{5 \sqrt{3}}{3}