Finding the affine equation of a plane given three points First of all, I am not sure the correct word is "affine" in english however it is "affin" in Swedish and I can't find the english counterpart. Affine form is basically the non-parametric equation of a plane: a
$$ax + by + cz + d = 0$$
I am given three points:
$$P:(1,2,0)$$ 
$$Q:(-1,3,1)$$ 
$$R:(-1,1,2)$$ 
and asked to find the "non-parametric" equation for the plane. 
My attempt is to simply solve the system of equations: 
$$ax + by + cz + d = 0$$
$$\begin{cases} 
a + 2b + d = 0
\\
-a + 3b + c + d = 0
\\
-a+b+2c+d = 0
\end{cases}$$
The process is quite tedious so I won't bore you with it, but I know it's correct and I end up with: 
$$
\begin{cases}
a = -2b-d
\\
b = ?
\\c=2b
\\
d=-\frac{7}{2}b
\end{cases}
$$
There is too many unknown variables too solve the system of equations, but in my book the author simply assigns $b = 2$ and solves it. Why can this be done? 
 A: The Cartesian equation of a plane in $\Bbb R^3$ can neatly be expressed as $\vec n\cdot\vec x=\vec n\cdot\vec x_0$ where $\vec n$ is a vector normal to the plane, $\vec x=\langle x,y,z\rangle$, and $\vec x_0$ is a point on the plane. In our case, the two directions
\begin{align*}
\vec{PQ} &= \langle -2,1,1\rangle & \vec{PR} &= \langle-2,-1,2\rangle
\end{align*}
are both in the plane. Therefore we may take 
$$
\vec n=\vec{PQ}\times\vec{PR}=\langle3,2,4\rangle
$$
We may also take $\vec x_0=P$. This gives 
$$
3\,x+2\,y+4\,z=7
$$
A: $$
P-Q = (2,-1,-1) \\
Q-R=(0,2,-1) \\
\hat n = (P-Q)\times (Q-R) =(3,2,4)
$$
now
$$
(\hat r - Q)\cdot\hat n = 0
$$
i.e.
$$
3x+2y+4z=7
$$
A: It is because the parameters that describe the plane are not unique. If $ax+by+cz+d=0$ then also $2ax+2by+2cz+2d=0$. Of course the $2$ there could be any non-zero real number. 
The work you did shows us that if $b=0$ then all of $a$, $c$ and $d$ would also be zero. This shows that in case of this particular plane $b$ cannot be zero. And by the earlier argument we can choose it arbitrarily.
A: Better to choose an equation of a plane with three constants like:
$$\dfrac{x}{a} + \dfrac{y}{b} +  \dfrac{z}{c}  = 1     $$ 
(  eliminating $d$ ), where  (a,b,c)  can be solved in the same way. 
Geometrically it means we are not leaving out minimum distance related to $d$ to the origin arbitrarily.   
