Show that the line `(x-1)/2=(y-2)/3=(z-3)/4a n d(x-4)/5=(y-1)/2` intersect. Find their point of intersection.

Explanation

We have `x_(1)=1,y_(1)=2,z_(1)=3` and `a_(1)=2,b_(1)=3,c_(1)=4`
Also `x_(2)=4,y_(2)=1,z_(2)=0` and `a_(2)=5,b_(2)=2,c_(2)=1`
If two lines intersect then shortest distance between them should be zero.
`:.` Shortest distance between two given lines
`=(|(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)-b_(2)c_(1))^(2)+(c_(1)a_(2)-c_(2)a_(1))^(2)+(a_(1)b_(2)-a_(2)b_(1))^(2))`
`(|(4-1,1-2,0-3),(2,3,4),(5,2,1)|)/(sqrt((3.1-2.4)^(2)+(4.5-1.2)^(2)+(2.2-5.3)^(2))`
`=(|(3,-1,-3),(2,3,4),(5,2,1)|)/(sqrt(25+34+121))`
`=(3(3-8)+1(2-20)-3(4-15))/(sqrt(470))`
`=(-15-18+33)/(sqrt(470))=0/(sqrt(470))=0`
Therefore the given two lines are intersecting.
For finding their point of intersection for the first line
`(x-1)/2=(y-2)/3=(z-3)/4=lamda`
`implies x=2lamda+1,y=3lamda+2` and `z=4lamda+3`
Since the lines are intersecting. So, let us put these values in the equation of another line.
Thus, `(2lamda+1-4)/5=(3lamda+2-1)/2=(4lamda+3)/1`
`implies (2lamda-3)/5=(3lamda+1)/2=(4lamda+3)/1`
`implies (2lamda-3)/5=(4lamda+3)/1`
`implies 2lamda-3=20lamda+15`
`implies 18lamda=-18=-1`
So the required point of intersection is
`x=2(-1)+1=-1`
`y=3(-1)+2=-1`
`z=4(-1)+3=-1`
Thus, the ines intersect at `(-1,-1,-1)`.