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