Find the Cartesian equation of the plane containing the following 2 lines

L1: x-1 = y/2 = (z+1)/-3
L2: (x+1)/-2 = y-1 , z = -1

So far i wrote the lines out into parametric form then into their equations:

L1 = i-k+t(i+2j-3k)
L2 = -i-j+k+t(-2i+j)

Then i used the cross product to get the normal vector:

n = 3i+6j+5k

The final answer is 3i+6j+5k = -2

I'm not sure where the -2 comes from, any help would be appreciated!!

I must disagree with your equation for L2. If the given conditions are correct, it should look something like this:

L1: = i - k + t(i + 2j - 3k)
L2: = -i + j + k + u(-2i + j)

Aside from changing one of the signs, I introduced u as a second parameter. Using the same parameter twice causes problems when the equations are combined.

From L2, we can se that the z-coordinate must be -1 for any point on the line. If it intersects L1 (and it had better), then the z-coordinate of the intersection point is -1. Use that to find t.

t = (z + 1)/3 = (-1 + 1)/3 = 0
x - 1 = 0
x = 1
y/2 = 0
y = 0

The intersection point would have to be (1, 0, -1). Check that against L2.

u = (x + 1)/(-2) = (1 + 1)/(-2) = -1
u = y - 1 = 0 - 1 = -1
It checks, so the lines do intersect at (1, 0, -1), and there does exist a plane that includes both lines.

You have already found a normal vector <3, 6, 5>, with which I concur. Use this general form for the plane:

3x + 6y + 5z + k = 0

To find k, substitute any point that falls on either line. We may as well use the intersection point from above.

3(1) + 6(0) + 5(-1) + k = 0
k = 2

3x + 6y + 5z + 2 = 0