For a line to be $\perp$ to a plane, $d.r.$ of the line should be same as that of normal to the plane.
From the equation of $L$, $d.r.$ of the line =$\overrightarrow b= (3,2,-1)$
From the equation of $P$, normal to the plane is $\overrightarrow n=(1,-2,-1)$
$\Rightarrow\:$the line is not $\perp$ to the plane.
But since $\overrightarrow b.\overrightarrow n=0$, the line can be || to the plane.
Since the point on the line $\overrightarrow a=(1,-1,3)$ satisfies the equation of the plane,
The line lies on the plane.