Answer
The lines are parallel.
Work Step by Step
Step 1
For two lines to be parallel, the vector from their parametric equations should be parallel. Vectors can be checked if their unit vectors are equal or conjugate with each other.
Step 2
Based from the form of parametric we equation form of
\begin{align*}
x&=x_0+at,\\
y&=y_0+bt,\\
z&=z_0+ct
\end{align*}
where $a$, $b$, and $c$ are the vector components parallel to the line we can determine the vector parallel to the given lines $\vec{L}_1$ and $\vec{L}_2$.
Get the vector $\vec{v}_1$ parallel to $\vec{L}_1$.
\begin{align*}
a_1&=-2,& b_1&=1,& c_1&=-1\\
\vec{v}_1&=\langle-2,1,-1\rangle
\end{align*}
Get the vector $\vec{v}_2$ parallel to $\vec{L}_2$.
\begin{align*}
a_2&=-4,& b_2&=2,& c_2&=-2\\
\vec{v}_2&=\langle-4,2,-2\rangle
\end{align*}
Step 3
Determine the unit vectors $\vec{v}_1$ and $\vec{v}_2$ by dividing the component by their magnitude. The magnitude formula is $\sqrt{x^2+y^2+z^2}$.
Get the unit vector $\vec{v}_1$.
\begin{align*}
|\vec{v}_1|&=\sqrt{(-2)^2+1^2+(-1)^2}= \sqrt{6}\\
\hat{v}_1&=\frac{\vec{v}_1}{|\vec{v}_1|}=\langle-2/\sqrt{6},1/\sqrt{6},-1/\sqrt{6}\rangle\\
&=\langle-\sqrt{6}/3,\sqrt{6}/3,-\sqrt{6}/3\rangle
\end{align*}
Get the unit vector $\vec{v}_2$.
\begin{align*}
|\vec{v}_2|&=\sqrt{(-4)^2+2^2+(-2)^2}=\sqrt{24}=2\sqrt{6}\\
\hat{v}_2&=\frac{\vec{v}_2}{|\vec{v}_2|}=\langle-4/2\sqrt{6},2/2\sqrt{6},-2/2\sqrt{6}\rangle\\
&=\langle-\sqrt{6}/3,\sqrt{6}/3,-\sqrt{6}/3\rangle
\end{align*}
Step 4
Since the vectors parallel to the lines have the same unit vector, the lines are parallel.
Result
The lines are parallel.
