Finding the equation of a plane that passes through two points and is parallel to a line 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?
 A: 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
$$
\cases{
(1,2,3)\cdot\vec n = 0\\
(2,-1,1)\cdot\vec n = 0\\
\|\vec n\|=1
}
$$
A: 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.
A: Follow these steps:


*

*The origin $\boldsymbol{r}_0$ and the point $ \boldsymbol{r}_A  $ define a line that belongs to the plane. This line has direction $(\boldsymbol{r}_A - \boldsymbol{r}_0)$. The normal to the plane must be perpendicular to all the vectors on this plane.

*Define the normal $\boldsymbol{n}$ as the vector that is perpendicular to the line $X$ and the line from the previous step. 
$$ \boldsymbol{n} = \boldsymbol{e} \times (\boldsymbol{r}_A - \boldsymbol{r}_0) $$
Here $\boldsymbol{e}$ is the direction vector of line $X$. Note that a simplified scalar multiple of $\boldsymbol{n}$ would do just fine.


*Find the distance $d$ of the plane to the origin. Use the equation of the plane $ \boldsymbol{n} \cdot \boldsymbol{r} = d$ that must be true for all points on the plane. Since the origin is one of the points, it stands to reason that $$ d = \boldsymbol{n} \cdot \boldsymbol{r}_0 = 0 $$

*The equation of the plane is thus found by
$$ \boldsymbol{n} \cdot \pmatrix{x\\y\\z} = 0 $$
Here $\cdot$ is the vector dot product, and $\times$ the vector cross product.
A: There are many different ways to solve this problem without cross product. The more straight could posssibly be the following.   
Let's call $P=(1,2,3)$.   
1) find the pencil of planes containg $O,P$
To do this, we just have to find two planes, not parallel to each other, and passing through those points. We can take for instance a plane parallel to the $z$ axis and passing through the projections of the points on $x,y$ plane: $\pi_1\; :\; 2x-y=0$ . Same way, a plane parallel to $y$ axis: $ \pi_2 \; : \; 3x-z=0$.
So the pencil is $0=\mu \pi_1 + \lambda \pi_2=(2\mu +3\lambda) x -\mu y - \lambda z$.
2)find a plane in the pencil, parallel to the given line
Just take two points on the line, the vector defined by the corresponding segment (in your case it is already given)and impose that it be normal to the normal of the pencil. 
You get a simple homogeneous equation in $\mu,\lambda$.  That is
$$
\eqalign{
  & 2\left( {2\mu  + 3\lambda } \right) - ( - 1)\mu  - \lambda  = 5\mu  + 5\lambda  = 0\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \mu  =  - \lambda \quad  \Rightarrow \quad \left\{ \matrix{
  \mu  = 1 \hfill \cr 
  \lambda  =  - 1 \hfill \cr}  \right. \cr} 
$$
Therefore the required plane is
$$
\pi :\quad  - x - y + z = 0
$$
which in fact contains $(0,0,0)$ and $(1,2,3)$ and is parallel to the line because
$(2,-1,1) \cdot (-1,-1,1)=0$.
This indicates another way to solve the problem, i.e. the system
$$
ax + by + cz = d\;:\quad \left( {\matrix{
   0 & 0 & 0  \cr 
   1 & 2 & 3  \cr 
   2 & { - 1} & 1  \cr 
 } } \right)\left( {\matrix{
   a  \cr 
   b  \cr 
   c  \cr 
 } } \right) = \left( {\matrix{
   d  \cr 
   d  \cr 
   0  \cr 
 } } \right)
$$
