3D Line Equations - Multivariable Calculus

Consider the line $L$ given by the symmetric equations $\frac{x-3}{2} = \frac{y+1}{-1} = \frac{z-5}{4}$ . Which of the following lines is parallel to $L$ and passes through the origin?
Answer: $\vec{r} = t\langle 2, -1, 4 \rangle$ For lines to be parallel, they must have direction vectors that are scalar multiples of each other. The given line L has direction vector ⟨2, -1, 4⟩ from the denominators in the symmetric equation. A line through the origin has the form ⟨0, 0, 0⟩ + t⟨a, b, c⟩ = t⟨a, b, c⟩. Choice A has direction vector ⟨2, -1, 4⟩, which is exactly the same as L's direction vector, so it's parallel and passes through the origin. Choice B has direction vector ⟨3, -1, 5⟩, which is not a scalar multiple of ⟨2, -1, 4⟩. Choice C has direction vector ⟨1, -2, 2⟩, which is also not a scalar multiple of ⟨2, -1, 4⟩. Choice D doesn't pass through the origin since when x = z = 0, we get y = -1, not y = 0.