I want to find a line in the plane denoted by points $P_0=(0,0,0)$,$P_1=(2,2,0)$,$P_2=(0,1,-2)$. The line I am looking for have to be perpendicular to line $${x+1\over3}={y-1\over2}={z\over{1\over2}}$$
What I have done so far:
$\vec n=\vec{P_0P_1} \times \vec{P_0P_2}=-4i-4j+2k$ is perpendicular to the plane
Then $\vec a=<3,2,{1\over2}>$ is in the direction of the given line.
The line I am searching for is of the form $\vec r=\vec x+t\vec b$. So I imagine I have to find $\vec x$ , $\vec b$
For $\vec b$ I know that $\vec b \perp \vec a$ whcih means $\vec b \cdot \vec a=0$ I also know that $\vec b \perp \vec n$ which means $\vec b \cdot \vec n=0$
Then If $\vec b=<b_1,b_2,b_3>$ I have two equations with 3 unknowns and I am missing one to be able to find $\vec b$.
I also don't know how to find $\vec x$
Any ideas how to proceed?
Firstly, the normal to the plane is actually $$\left(\begin{matrix}-4\\4\\2\end{matrix}\right)$$ Then you want the direction of the required line to be perpendicular to this normal (i.e. parallel to the plane) and perpendicular to the line given with direction vector $$\left(\begin{matrix}3\\2\\\frac 12\end{matrix}\right)$$
Therefore work out the cross-product of these vectors to get the direction. However you need to be specific about what point you would like this line to pass through, since there are infinitely many possible choices.