Answer
$\left\{\begin{array}{l}
x=1-t\\
y=3t\\
z=-1+t
\end{array}\right., \quad 0\leq t \leq 1$
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
$\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
---
$\overrightarrow{PQ}=\langle 0-1,3-0, 0+1 \rangle=\langle-1,3,1\rangle={\bf v}$.
A point on the line is $P(1,0,-1)$, so a parametrization can be
$\left\{\begin{array}{l}
x=1-t\\
y=0+3t\\
z=-1+t
\end{array}\right., \quad -\infty \lt t \lt \infty$
when $t=0$, the point defined is $(2,0,2)$
when $t=1$, the point defined is $(0,2,0)$
so, the line segment is parametrized with
$\left\{\begin{array}{l}
x=1-t\\
y=3t\\
z=-1+t
\end{array}\right., \quad 0\leq t \leq 1$
