Find the algebraic equation of the plane that passes through the origin, is parallel to the line $\mathbf X= (-1, 0, 2) +t(2,-1,1)$, and contains the point $P=(1,2,3)$.
I suspect you can do this with a cross-product, but our professor isn't letting us use that as he hasn't taught it to us yet. Is there a way to get another point on the plane from the parallel line, so you can make a system of equations to solve for an orthogonal vector?
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Given the points $\{p_0,p_1\}\in \Pi$ and the line $L\to p = p_2+\lambda \vec v$, determine a plane $\Pi$ parallel to $L$.
Characterizing the plane as $\Pi\to(p-p_0)\cdot\vec n=0$ we have the properties
$$ \cases{ (p_1-p_0)\cdot \vec n = 0\\ \vec v\cdot \vec n = 0\\ \|\vec n\|=1 } $$
The last comes from normalization purposes.
Now as
$$ \cases{ p=(x,y,z)\\ p_0=(0,0,0)\\ p_1 = (1,2,3)\\ \vec v = (2,-1,1) } $$
we have to solve
$$ \cases{ (1,2,3)\cdot\vec n = 0\\ (2,-1,1)\cdot\vec n = 0\\ \|\vec n\|=1 } $$
The equation of a plane is usually of the form $ax+by+cz+d=0$. If the origin lies in the plane (as it does in the question), then $d=0$ and one may write the equation as $\langle (a,b,c),(x,y,z)\rangle=0$.
Now, we know $(1,2,3)$ lies in the plane. Moreover, for all $t\in \mathbb{R}$, $t\cdot(2,-1,1)$ also lies in the plane. Hence, $(a,b,c)$ must satisfy
$$a+2b+3c=0\\2a-b+c=0$$
You probably see this is a system of two equations with $3$ unknowns, so the solution is not unique. That's okay, because for any $\alpha \in \mathbb{R}\setminus\{0\}$ the equations $ax+by+cz=0$ and $\alpha\cdot(ax+by+cz)=0$ are the same.