Do l1 : (3,2,3,-1) + t(1,6,1,-1) and l2 : (1,0,1,1) + t(1,-3,-4,2), where TER intersect in R^4 and where?

Answer 1

I think you want different parameters for I1 and I2; otherwise the answer is clearly no, since if you try to match the first component of I1 with the first component of I2 you get
3+t = 1+t, which has no solution.

So instead let l1 : (3,2,3,-1) + t(1,6,1,-1) and l2 : (1,0,1,1) + s(1,-3,-4,2) where s and t are real parameters. An intersection would mean that the following must be true simultaneously:

3 + t = 1 + s
2 + 6t = -3s
3 + t = 1 - 4s
-1 - t = 1 + 2s

If you solve the first and third equations simultaneously, you will get s = 0, t = -2.These values satisfy the fourth equation, but do not satisfy the second equation. So there is no intersection.