Rewrite the lines as:
L1:(1,2,0)+t(-6,9,-3)
L2:(2,3,0)+u(2,-3,1)
The direction vector of L1 is <-6,9,-3>=-3<2,-3,1>
while that of L2 is <2,-3,1>
we see that both direction vectors are parallel, that is, L1 and L2 are parallel.
To check if the two lines are coincident, we need to find at least one common point.
Assuming the x-coordinates coincide, we have
1-6t=2+2u => 6t+2u=-1 ................(1)
Assuming the y-cordinates coincide, we have
2+9t=3-3u => 9t+3u=1..................(2)
3(1)-2(3) : 18t+6u - (18t+6u) = -2-3 => 0=-5 ..... therefore no solution.
If we cannot find one common point between the two lines, they are not coincident.
Another way to check this is to match the x-coordinates of the lines, and see if the other coordinates match
Here, we try to match the x-coordinate = 1, for the point (1,2,0) on L1.
For L2, we set u=-1/2
L2: (2,3,0)-(2,-3,1)/2 = (2-1, 3+1.5, 0-0.5) = (1, 4.5, -0.5)
which does not match (1,2,0) on L1. So the two lines are not coincident.
Answer: the two lines L1 and L2 are parallel (i.e. not coincident)
