Answer
Parametrization of the line is ${\bf{r}}\left( t \right) = \left( {4 + t,9,8} \right)$
Work Step by Step
A line perpendicular to the $yz$-plane has direction parallel to the $x$-axis. So, the direction vector is ${\bf{v}} = {\bf{i}}$, where ${\bf{i}} = \left( {1,0,0} \right)$. Since, it passes through the point $P = \left( {4,9,8} \right)$, we write ${{\bf{r}}_0} = \overrightarrow {OP} = \left( {4,9,8} \right)$.
Using Eq. (5), the line passes through $P = \left( {4,9,8} \right)$ in the direction of ${\bf{v}} = \left( {1,0,0} \right)$ has vector parametrization:
${\bf{r}}\left( t \right) = {{\bf{r}}_0} + t{\bf{v}} = \left( {4,9,8} \right) + t\left( {1,0,0} \right)$
${\bf{r}}\left( t \right) = \left( {4 + t,9,8} \right)$
So, the parametric equations are: $x=4+t$, $y=9$, $z=8$, where $ - \infty < t < \infty $.
