Find the shortest distance between the two lines whose vector equations are given by: `vecr=(1+lamda)hati+(2-lamda)hatj+(-1+lamda)hatk and vecr=2(1+mu

The lines are `vecr=(1-lamda)hati+(lamdal-2)hatj+(3-2lamda)hatk`
and `vecr=(mu+1)hati+(2mu-1)hatj-(2mu+1)hatk`
or `" "vecr=(hati-2hatj+3hatk)+lamda(-hati-2hatj-2hatk)`
and `" "vecr=(hati-hatj-hatk)+mu(hati+2hatj-2hatk)`.
Line (i) passes through the point `(x_(1), y_(1), z_(1))-=(1, -2, 3)` and is parallel to the vector
`" "a_(1)hati+b_(1)hatj+c_(1)hatk-=-hati-2hatj-2hatk`.
Line (ii) passes through the point `(x_(2), y_(2), z_(2))-= (1, -1, -1)` and is parallel to the vector
`" "a_(2)hati+b_(2)hatj+c_(2)hatk-= hati+2hatj-2hatk`.
Hence, the shortest distance between the lines using the formulla
`(||{:(x_(2)-x_(1),,y_(2)-y_(1),,z_(2)-z_(1)),(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)):}||)/(||{:(hati,,hatj,,hatk),(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)):}||)` is
`(||{:(1-1,,-1+2,,-1-3),(-1,,-2,,-2),(1,,2,,-2):}||)/(||{:(hati,,hatj,,hatk),(-1,,-2,,-2),(1,,2,,-2):}||)=(4)/(sqrt80)=(1)/(sqrt(5))`