Find the length and the equations of the line of shortest distance between the lines
`(x+1)/(7)=(y+1)/(-6) =(z+1)/(1) " and " (x-3)/(1)=(y-5)/(-2) =(z-7)/(1)`

Explanation

The equations of the given lines are
`(x+1)/(7) =(y+1)/(-6) =(z+1)/(1)=lambda` (say)
`(x-3)/(1) =(y-5)/(-2) =(z-7)/(1)=mu `(say)
Any point on (i) is `P(7lambda -1, -6lambda -1,lambda -1)`
Any point on (ii) is `Q(mu +3 ,-2mu+ 5,mu +7)`
The direction rations of PQ are `(mu-7 lambda+4 ,-2mu +6lambda +6 ,mu -lambda +8)`
Now PQ will be the shortest distance between (i) and (ii) only
when PQ is perpendicular to both (i) and (ii)
`:. {underset( 1.(mu -7 lambda +4) -2(-2mu +6lambda+6)+1.(mu -lambda +8) =0)underset( " and ")(7(mu -7lambda +4)-6(-2mu +6lambda +6) +1.(mu-lambda +8)=0)`
`rArr {underset(6 mu -20lambda =0 )(20 mu -86 lambda=0) rArr {underset(3 mu -10lambda)(10mu -43lambda) rArr lambda=0 " and " mu =0`
`:. ` PQ will be the line of shortest distance when `lambda =0 "and " mu=0`
Putting `lambda=0 " and " mu =0` we get the points `P(-1,-1,-1)` and Q(3,5,7)
`:. SD =|PQ|=sqrt((3+1)^(2)+(5+1)^(2)+(7+1)^(2))`
`=sqrt((4)^(2) +(6)^(2)+(8)^(2))=(sqrt(16+36+64)`
`=sqrt(116) =2 sqrt(29) ` units
Clearly the equations of the line of the shortest distance is the equation of line PQ given by
`(x-3)/(3+1) =(y-5)/(5+1) =(z-7)/(7+1)`
`hArr (x-3)/(4)=(y-5)/(6) =(z-7)/(8)`
`hArr (x-3)/(2)=(y-5)/(3)=(z-7)/(4)`
Hence the equations of the line of shortest distance is
`(x-3)/(2)=(y-5)/(3)=(z-7)/(4)`