Answer
$\left\{\begin{array}{l}
x=-2+t\\
y=t\\
z=3-t
\end{array}\right., \quad -\infty \lt t \lt \infty$
(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),
$\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
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Here, $\overrightarrow{PQ}=\langle 3+2,5-0,-2-3 \rangle=\langle 5,5, -5 \rangle=5\langle 1,1, -1 \rangle$
so we take ${\bf v}$=$\langle 1,1, -1 \rangle$.
A point on the line is $P(-2,0,3)$, so a parametrization can be
$\left\{\begin{array}{l}
x=-2+t\\
y=t\\
z=3-t
\end{array}\right., \quad -\infty \lt t \lt \infty$
