The shortest distance between the lines $$x - y = 0 = 2x + z$$     and $$x + y - 2 = 0 = 3x - y + z - 1$$       is :

Solution :

Equations of the lines in symmetric form are $$\frac{x}{1} = \frac{y}{1} = \frac{z}{{ - 2}}{\text{ and }}\frac{{x - \frac{3}{4}}}{1} = \frac{{y - \frac{5}{4}}}{{ - 1}} = \frac{z}{{ - 4}}$$
$$\eqalign{ & {\text{Now,}}\,\,l + m - 2n = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,l - m - 4n = 0 \cr & {\text{Solving, }}\frac{l}{{ - 2}} = \frac{m}{2} = \frac{n}{{ - 2}}\,\,\,\,\,\, \Rightarrow l = \frac{{ - 1}}{{\sqrt 3 }},\,m = \frac{1}{{\sqrt 3 }},\,n = \frac{{ - 1}}{{\sqrt 3 }} \cr} $$
$$\therefore $$  shortest distance $$ = \left( {\frac{3}{4} - 0} \right)\frac{{ - 1}}{{\sqrt 3 }} + \left( {\frac{5}{4} - 0} \right)\frac{1}{{\sqrt 3 }} + \left( {0 - 0} \right)\frac{{ - 1}}{{\sqrt 3 }}.$$