<span>The lines are skew.
I am assuming that there's a typo in the equation for the l1. The equation t = 2 + 6t should be z = 2 + 6t.
So the two lines are:
l1: x =6 + 4t, y = 8 - 2t, z = 2 + 6t
l2: x = 4+16s, y = 12 - 8s, z = 16 + 20s
Let's check if the lines are parallel.
Direction vector for l1 is <4,-2,6) and l2 is <16,-8,20>. At first glance the direction vector for l2 looks to be 4 times that of l1, let's divide l2's vector by 4 and verify that it's proportional to l1's vector. <16/4, -8/4, 20/4> = <4, -2, 5>. So l1 and l2 are NOT parallel since their direction vectors are not proportional.
Let's see if the lines intersect. Set their equations equal to each other.
(1) 6 + 4t = 4+16s
(2) 8 - 2t = 12 - 8s
(3) 2 + 6t = 16 + 20s
We have 2 unknowns and 3 equations. Let's triple equation (2) above and add to equation (3) in order to solve for s. Note: Equations (2) and (3) were selected since the ratios in the direct vectors for the x and y components are the same. So selecting (1) and (2) would cancel out both s & t. So (3) has to be used as one of the equations and either (1) or (2) can be used. But the ratios for (2) are simpler than for (1). So (2) and (3) are the equations selected.
24 - 6t = 36 - 24s
2 + 6t = 16 + 20s
26 = 52 - 4s
-26 = -4s
6.5 = s
2 + 6t = 16 + 20s
2 + 6t = 16 + 20*6.5
2 + 6t = 16 + 130
2 + 6t = 146
6t = 144
t = 24
So we have a s of 6.5 and a t of 24. Let's check that the points match.
l1: x =6 + 4t, y = 8 - 2t, z = 2 + 6t
l1: x =6 + 4*24, y = 8 - 2*24, z = 2 + 6*24
l1: x =6 + 96, y = 8 - 48, z = 2 + 144
l1: x =102, y = -40, z = 146
l2: x = 4+16s, y = 12 - 8s, z = 16 + 20s
l2: x = 4+16*6.5, y = 12 - 8*6.5, z = 16 + 20*6.5
l2: x = 4+104, y = 12 - 52, z = 16 + 130
l2: x = 108, y = -40, z = 146
If we get the y and z coordinates to match, the x coordinates don't match, so the lines don't intersect.
And since the lines are not parallel, nor do they intersect, then they must be skew.</span>
