Find the parametric equation of a line different from, yet parallel to $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$

Question

Line $L_2$ is $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$

Find the parametric equation of a line $L_4$ which is different from, yet parallel to, the line $L_2$ given above.

Where I am at so far:

All I know is that two lines in three dimensions are parallel if the direction vectors of both lines are scalar multiples of each other. So I know $L_4$'s direction vector is $(1, -2, -1)$. But that's all I got.

Answer 1

Convince yourself that $(0,0,0)$ is not a point on $L_2$.

Then

$$L_4: \quad ( x, y, z ) = (0, 0, 0) + t(1, -2, -1)$$

will do the job.

Answer 2

In the equation $$( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$$ the $(1, 1, -1)$ is a point on the line (for $t=0$) and the $(1, -2, -1)$ is a vector, that gives the direction of the line. Therefore parallel lines are $$( x, y, z ) = (\alpha, \beta, \gamma) + t(1, -2, -1)$$ and $\alpha, \beta, \gamma\in \mathbb{R}$.