find the line through P that is perpendicular to the plane in part a

Question

Let $P=(1,1,0), Q=(2,0,-1), R=(2,-1,1)$ be 3 points in $\mathbb{R}^3$.

a) find the plane containing $P,Q$, and $R$. (I know how to do this part)

$-3(x-1)-2(y-1)-1(z-0)=0$ ?

b) find the line through P that is perpendicular to the plane in part a).

I think I might know how to do this, but I'm not sure. Am I supposed to take the vector normal to the plane (The coefficients of the plane equation)$(\{-3,-2,-1\})$, multiply that by $t$, and add it to the vector version of $P$ ? $\{1,1,0\}$

this was on a quiz, so I don't have the answer off hand

Answer 1

The parametric equations of the line will be $$x=1-3t$$ $$y=1-2t $$ $$z=-t $$

the cartesian are

$$x-3z=1$$ and $$y-2z=1$$

which represent two planes.