cal3 intersection of lines, help?

Question

given two lines
L1: x-2=(y+1)/2=(z-3)/3=t
L2:(x-5)/3=(y-1)/2=z-4=s

can any one show me how to solve for the point of intersection?

Answer 1

L1: x - 2 = (y + 1)/2 = (z-3)/3 = t
L2: (x - 5)/3 = (y - 1)/2 = z - 4 = s

Can any one show me how to solve for the point P, of intersection?
________

Solve for x, y, and z. Get the parametric equations of the lines.

L1:
x = 2 + t
y = -1 + 2t
z = 3 + 3t

L2:
x = 5 + 3s
y = 1 + 2s
z = 4 + s

Solve for s and t.

x: 2 + t = 5 + 3s
y: -1 + 2t = 1 + 2s
z: 3 + 3t = 4 + s

x: t = 3 + 3s
y: 2t = 2 + 2s
z: 3t = 1 + s

Plug the value of t from x into y.
y: 2(3 + 3s) = 2 + 2s
6 + 6s = 2 + 2s
4s = -4
s = -1

Plug into x.
t = 3 + 3(-1) = 0

Check to make sure this relationship holds for z.
z: 3t = 1 + s
3*0 = 1 - 1*(-1)
0 = 0

It does so the lines intersect.

Calculate the point P of intersection.

x = 2 + t = 2 + 0 = 2
y = -1 + 2t = -1 + 0 = -1
z = 3 + 3t = 3 + 0 = 3

The point of intersection is P(2, -1, 3).

Answer 2

I believe that each x,y, and z point has to be computed independently of each other. For example, the z coordinate will be such that z-4 =
(z-3)/3.