L1 =(4+2t)i + (3-2t)j + (-7-4t)k
L2 =(t1)i + (3+3t1)j + (1-2t1)k
L2 =(t1)i + (3+3t1)j + (1-2t1)k
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If they do intersect, there will be a pair of values for the parameters t and t1 that product the same x, y, z triple. I'm gonna call the second parameter s---it's easier to type.
Set them equal
4 + 2t = s,
3 - 2t = 3 + 3s, and
-7 - 4t = 1 - 2s.
From equations (1) and (2)
s = 2t + 4, and 3s = -2t ==> 3(2t + 4) = -2t ==> 12 = -8t ==> t = -3/2 and s = 1.
If these solve equation (3), then the lines intersect. We can check.
-7 - 4t = 1 - 2s, sub in t = -3/2 and s = 1
-7 -4(-3/2) = -1, and 1 - 2(1) = -1.
It checks out. So the lines intersect when t = -3/2 and s = 1. The triple (x, y, z) is obtained by substitution into either line. Using L2
x = s, y = 3 + 3s, z = 1 - 2s ==> when s = 1 (x, y, z) = (1, 6, -1).
Set them equal
4 + 2t = s,
3 - 2t = 3 + 3s, and
-7 - 4t = 1 - 2s.
From equations (1) and (2)
s = 2t + 4, and 3s = -2t ==> 3(2t + 4) = -2t ==> 12 = -8t ==> t = -3/2 and s = 1.
If these solve equation (3), then the lines intersect. We can check.
-7 - 4t = 1 - 2s, sub in t = -3/2 and s = 1
-7 -4(-3/2) = -1, and 1 - 2(1) = -1.
It checks out. So the lines intersect when t = -3/2 and s = 1. The triple (x, y, z) is obtained by substitution into either line. Using L2
x = s, y = 3 + 3s, z = 1 - 2s ==> when s = 1 (x, y, z) = (1, 6, -1).