Answer:
Step-by-step explanation:
Given are two lines equation in parametric form as
![l1: x=3-2t, y=7+4t, z=-3+8t\\l2: x=-1-u, y=18+3u, z=7+2u](https://tex.z-dn.net/?f=l1%3A%20x%3D3-2t%2C%20y%3D7%2B4t%2C%20z%3D-3%2B8t%5C%5Cl2%3A%20x%3D-1-u%2C%20y%3D18%2B3u%2C%20z%3D7%2B2u)
The direction ratios of the I line are -2, 4 and 8
Ii line are -1, 3, 2
These two are not proportional
Hence l1 and l2 are not parallel
Either they are skew or intersect
If intersect common point P would have coordinates as
![3-2t =- 1-u : 2t-u =4 \\7+4t = 18+3u: 4t-3u = 11\\-3+8t = 7+2u\\8t-2u = 10](https://tex.z-dn.net/?f=3-2t%20%3D-%201-u%20%20%3A%20%202t-u%20%3D4%20%5C%5C7%2B4t%20%3D%2018%2B3u%3A%20%20%204t-3u%20%3D%2011%5C%5C-3%2B8t%20%3D%207%2B2u%5C%5C8t-2u%20%3D%2010)
Let us solve first two equatins for u and t and check whether 3rd equation is satisfied by this.
2*(i) -ii gives u = -3
t =0.5
Substitute in III equation
8(0.5)-2(-3) = 10 so satisifed
These two are not skew lines
The point of intersection is got by substituting either t or u
Point of intersection (2, 9, 1)