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Let pi be a plane that contains the line $l$ with the parameter form $(x,y,z)=(2,-1,3)+t(1,2,1)$. The plane contains the point $P=(1,1,0)$. Determine the equation of the plane and the parameter form for a line through the point $P$ that is lying on the plane pi and that is perpendicular to $l$.

I'm not sure about this question, but shall I first determine a line's equation on the form x=x0+tv, where x0 is the point $P$ and v is the vector perpendicular to the line $l$? I don't know, it's just a feeling that I have inside me. I would appreciate some guiding with thinking around this problem :P

user2566415
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The equation for your plane $\pi$ will simply be of the form $\mathbf x = \mathbf x_0 + s\mathbf u + t\mathbf v$ (you need $2$ variables to specify a plane and only $1$ to specify a line). Given your explicit formula for $l$, you've already got $\mathbf x_0$ and $\mathbf u$.

You can find the vector $\mathbf v$ by simply taking any point on $l$ (which is in $\pi$) and subtracting it from the vector $\vec {OP}$. This will give another vector which is in $\pi$ as well. You'll want to make sure that $\mathbf v$ and $\mathbf u$ are not collinear.

Then to find the line, $l_2$, going through $P$ that is orthogonal to $l$ you again need an $\mathbf x_2$ and a $\mathbf u_2$. The $\mathbf x_2$ is supplied by $\vec {OP}$, though.

As for the direction vector, you should draw a picture to verify this:
$\mathbf u_2 = [(2,1,3) - (1,1,0)] - \dfrac{[(2,1,3) - (1,1,0)]\cdot (1,2,1)}{\|(1,2,1)\|^2}(1,2,1)$

What this equation is saying is that the vector pointing from $P$ to a particular point on your line $l$ (I chose the obvious one $(2,1,3)$) minus the projection of that vector onto a vector in the same direction as $l$ (the direction vector in the equation you have for $l$ is a good choice here) is a vector orthogonal to $l$, but still in the plane.