You have here a line defined in three dimensions, x, y, and z.
Your question involves parametric equations. You have the parameter t.
The z-coordinate is just z, no matter what the parameter t may be. This shows that the three dimensional line is in the plane z no matter what the parameter. Think of the z axis as a bunch of x-y axes all stacked and going back into the page (as you draw it on paper).
Since the parameter t does not effect the z-coordinate, you can think of this as a line that remains in a vertical x-y plane.
Then you can choose values for t that will yield values for x and y that can be plotted on that vertical x-y plane.
t x y
0 2 5 (2,5)
1 3 2 (3,2)
2 4 -1 (4,-1)
3 5 -4 (5,-4)
You then see the line has a slope of -3.
In that plane (z), a line parallel would also have a slope of -3.
You will recognize the -3 in the portion of the parametric equation for y: y = 5 -3t
To get a parallel line, simply change the 5 in the y-equation.
For example, y = 6 - 3t.
Keep the x and z equations the same to define the line in three dimensions.
(L2) {x=2+t,y=6-3t,z}
One problem with my answer: I do not see how the = -1 + 2t is relevant, unless it is meant to somehow define which z-plane you are in.
