Answer
$\left\{\begin{array}{l}
x=2+2t\\
y=-4-t\\
z=7+3t
\end{array}\right., \qquad \left\{\begin{array}{l}
x=-2-t\\
y=-2+\frac{1}{2}t\\
z=1-\frac{3}{2}t
\end{array}\right., \quad $
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\qquad $(Eq.3)
---
If ${\bf v}=\langle 2,-1,3\rangle, $ and a point on the line is $P(2,-4,7)$, a parametrization can be
$\left\{\begin{array}{l}
x=2+2t\\
y=-4-t\\
z=7+3t
\end{array}\right., \quad -\infty \lt t \lt \infty$
If ${\bf v}=\displaystyle \langle-1,-\frac{1}{2},-\frac{3}{2}\rangle, $ and a point on the line is $P(-2,-2,1)$, a parametrization can be
$\left\{\begin{array}{l}
x=-2-t\\
y=-2+\frac{1}{2}t\\
z=1-\frac{3}{2}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
