Find a vector equation and parametric equations for the line. The

The line passing through the point (1,0,8) and perpendicular to the plane ${\displaystyle x+3y+z=3}$
Let ${\displaystyle r_0=i+0j+8k}$
A line perpendicular to the plane ${\displaystyle x+3y+z=3}$ has the same direction as a normal vector to the plane, which is ${\displaystyle n=<1,3,1>}$
So, take ${\displaystyle v=i+3j+k}$
By using the vector eqaution for the line ${\displaystyle r=r_0+tv}$
${\displaystyle r=(i+0j+8k)+t(i+3j+k)}$
${\displaystyle =(1+t)i+(0+3t)j+(8+t)k}$
${\displaystyle =(1+t)i\left(3t\right)j+(8+t)k}$
Thus, the vector equation of the line is ${\displaystyle r\left(t\right)=(1+t)+i\left(3t\right)j+(8+t)k}$
Obtain the parametric equation of line as follows.
Let ${\displaystyle P_0=(x_0,y_0,z_0)=(1,0,8)}$ and ${\displaystyle v=<a,b,c\ge <1,3,1>}$
By the parametric equation for the line through the point ${\displaystyle (x_0,y_0,z_0)}$ and parallel to the direction vector ${\displaystyle <a,b,c>}$ are ${\displaystyle x=x_0+at,y=y_0+bt,z=z_0+ct}$
That implies,
${\displaystyle x=1+t,y=0+3t,z=8+t}$
${\displaystyle x=1+t,y=3t,z=8+t}$
Thus, the parametric equation of the line is
${\displaystyle x\left(t\right)=1+t,y\left(t\right)=3t,z\left(t\right)=8+t}$