Find the shortest distance between the lines `(x-1)/2=(y-2)/3=(z-3)/4a n d(x-2)/3=(y-4)/4=(z-5)/5` .

Equations of the lines are
`(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-4)/(4)=(z-5)/(5)`
Here, `x_(1)=1,y_(1)=2,z_(1)=3`
`x_(2)=2,y_(2)=4,z_(2)=5`
`a_(1)=2,b_(1)=3,c_(1)=4`
`a_(2)=3,b_(2)=4,c_(2)=5`
Shortet distance betweent the lines is
`({:(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):})/(sqrt((b_(1)c_(2)-c_(1)b_(2))+(c_(1)a_(2)-c_(2)a_(1))^(2)+(a_(1)b_(2)-a_(2)b_(1))^(2)))`
`=(|{:(1,2,2),(2,3,4),(3,4,5):}|)/(sqrt((15-16)^(2)+(12-10)^(2)+(8-9)^(2)))`
`=(1(15-16)-2(10-12)+2(8-9))/(sqrt(1+4+1))`
`=(-1+4-2)/(sqrt(6))=(1)/(sqrt(6))` units.