Answer
$\left\{\begin{array}{l}
x=1-t\\
y=2-t\\
z=-1+t
\end{array}\right., \quad -\infty \lt t \lt \infty$
(Other answers are possible.)
Work Step by Step
Given a point on the line $P(x_{0},y_{0},z_{0})$ and if the line is parallel to ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle$,
the standard parametrization is given by formula (6),
$\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
---
Here, $\overrightarrow{PQ}=\langle-1-1,0-2,1-(-1) \rangle=\langle-2,-2, 2 \rangle=2\langle-1,-1, 1 \rangle$
so we take ${\bf v}$=$\langle-1,-1, 1 \rangle$.
A point on the line is $P(1,2,-1)$, so a parametrization can be
$\left\{\begin{array}{l}
x=1-t\\
y=2-t\\
z=-1+t
\end{array}\right., \quad -\infty \lt t \lt \infty$
