Answer
See proof
Work Step by Step
$\vec{v_1}=\langle 3,1,2 \rangle$ is the parallel vector for line $L_1$.
$\vec{v_2}=\langle -6,-2,-4 \rangle$ is the parallel vector for line $L_2$.
We can see that $\vec{v_1}$ and $\vec{v_2}$ are parallel to each other.
For $t=0$ on $L_1$, we have the point $\langle 1,-2,0 \rangle$.
For $t=\frac{1}{2}$ on $L_2$, we have the point $\langle 1,-2,0 \rangle$.
Since $L_1$ and $L_2$ have parallel vectors and a common point, they are the same line.
Therefore, we can express the line as:
$$\boxed{L:\langle x,y,z \rangle = \langle 1,-2,0 \rangle + t\langle 3,1,2 \rangle}$$
