Planes equation

Question

I have two parallel lines:

T2: (x−3)/2 = (y-2)/1 = (z+2)/2;

T3: x = 1 + 4s, y = 1 + 2s, z = −3 + 4s, −∞ < s < ∞

And i need to find a plane that contains them both but the plane i keep getting only contains one of them when i draw them in matlab. My calculating steps:

1. n = s x s

n = (0;-16;0)

2. a(x−x0)+b(y−y0)+c(z−z0)=0

-16y+32=0

3. Ax+By+Cz+D=0

D = 32

4. -16y+64

y-4

Answer 1

The directions of the lines are $(2,1,2)$ and $(4,2,4)$. These are parallel so their cross product should be zero, not $(0,-16,0)$.
The zero vector gives the equation $0x+0y+0z=0$, which is true but not useful.
You need a second vector in the plane, not parallel to $(2,1,2)$. That would be a vector going from one line to the other.
Find a point on each line. The vector from one to the other won't be parallel to $(2,1,2)$