Answer
$\left\{\begin{array}{l}
x=t\\
y=t\\
z=1.5t
\end{array}\right., \quad 0\leq t \leq 1$
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 formula (6),
$\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
---
$\displaystyle \overrightarrow{PQ}=\langle 1-0,1-0, \frac{3}{2}-0 \rangle=\langle 1,1,\frac{3}{2}\rangle= {\bf v}$.
A point on the line is $P(0,0,0)$, so a parametrization can be
$\left\{\begin{array}{l}
x=0+t\\
y=0+t\\
z=0+1.5t
\end{array}\right., \quad -\infty \lt t \lt \infty$
when $t=0$, the point defined is $(0,0,0)$
when $t=$, the point defined is $(1,1,1.5)$
so, the line segment is parametrized with
$\left\{\begin{array}{l}
x=t\\
y=t\\
z=1.5t
\end{array}\right., \quad 0\leq t \leq 1$
