Considering the lines are skew and the values of (s) and (t) did not satisfy the parametric equations. Isn't that self explanatory? Meaning there is no intersection of the lines? Or could the lines be skew and still intersect?
Anyways, It's always awesome to learn a new way to verify math! So I'm going to give it a shot.
Attempt at verifying the solution:
##\vec {l_1}## = (6,−1,0)+t(3,1,−4)
= (6,−1,0)+(-1/4)(3,1,−4)
= (6,-1,0)+(-3/4, -1/4, 1)
=(21/4, -5/4, 1 )
##\vec {l_2}## = (4,0,5)+s(−1,1,5)
=(4,0,5)+(-5/4)(−1,1,5)
=(4,0,5)+(5/4,-5/4,-25/4)
=(5, -5/4, -45/4)
I guess they do intersect at (x=-5/4)
That's cool. I guess the lines can be skew and still intersect. Good to know, my textbook should explain that.
And I have applied your teachings Sensei =) by using LaTex and finding the point of intersection :)