Find the components of all vectors v which have length 2 and are perpendicular to both the lines $x = 4 + 3t, y = 2 − t, z = 1 + 5t$ and $x − y + z = 2$, $3x + 2y − 4z = 6$.
So far I know that the vector $(1, -1, 1)$ is normal to the line $x-y+z=2$, and so is $<3, 2, -4>$ to the line $3x+2y-4z=6$. So I took their cross products, and that gives me the vector perpendicular to those lines.
For the line given in parametric form, I'm starting to get really confused - don't I have to substitute those into one of the equations i.e $(4+3t) - (2-t) + (1+5t) = 2$? I'm doing that and whatever value I'm getting for t does not satisfy the equation. I'm not sure if I'm in the right track or not. :(