determine point of intersection of the given lines, if an intersection does exist

Question

L1: (x,y,z) = (4,-3,15) + t(6,-3,12) L2: (x,y,z) = (-2,1,-4) + s(-2,1-4)

I was able to get this
4+6t=-2-2s
-3-3t=1+s
15+12t=-4-4s

but then i got stuck because i couldnt get anything other than s=-2-3t

Answer 1

first of all, notice that the direction numbers of the first line are
a scalar multiple of the 2nd line, that is
(6,-3, 12) = -3(-2, 1, -4) , so we know they are parallel.
pretend we don't notice that and we blindly go ahead to solve.

we want the x of the first equation to be equal of the second
the y of the first to be equal of the 2nd, etc

4+6t = -2-2s
2s + 6t = -6
s + 3t = -3

-3 - 3t = 1+s
s + 3t = -4
subtract them
0 = 1
which is a contradiction, so they cannot intersect