How many planes are there with the desired property? Two points $A$ and $B$ are given, for example $A(2/5/7)$ and $B(4/11/16)$.
The object is to find all planes containing the points $A$ and $B$ with distance
$2$ to the origin.
I tried the hesse-normal-form, but since only one direction-vector is given,
the normal-vector cannot be calculated.
I also tried to add a third point $C(c_1,c_2,c_3)$ and set up the plane
equation, but this also led to nowhere.
Can anyone determine the desired planes and, for each one, an equation ?
 A: Let $C=\left(u,v,w\right)\in\mathbb{R}^{3}$ with $\left\Vert C\right\Vert =2$. 
Let $P$ be a plane with $A,B,C\in P$ and $C\perp P$. 
Then $P$ is a plane with $d\left(P,\left(0,0,0\right)\right)=\left\Vert C\right\Vert =2$
and we have $3$ equations in $u,v,w$:
$\left(C,C\right)=4$
$\left(C,A-C\right)=0$ or equivalently $\left(C,A\right)=\left(C,C\right)=4$
$\left(C,B-C\right)=0$ or equivalently $\left(C,B\right)=\left(C,C\right)=4$
Can you solve this?
A: Let plane $\alpha$ be $p(X)=Ax+By+Cz+D=0$, then $\rho(\alpha,X)=\frac{|p(X)|}{\sqrt{A^2+B^2+C^2}}$.
So we put $A^2+B^2+C^2=1$ and get equations
$D=\pm 2$;
$2A+5B+7C+D=0$;
$4A+11B+16C+D=0$.
$2A+5B+7C+D=0 \Leftrightarrow A=-\frac{5}{2}B-\frac{7}{2}C\mp1$
Plugging this into $4A+11B+16C\pm 2=0$ we get 
$-10B-14C\mp 4 +11B+16C\pm 2 =0$.
$B+2C\mp 2=0$, $B=-2(C\mp 1)$, plug it back into $A=-\frac{5}{2}B-\frac{7}{2}C\mp1$ we get $A=5(C\mp 1)-\frac{7}{2}C\mp1=\frac{3}{2}C\mp 6$, and plug them all into $A^2+B^2+C^2=1$:
$(\frac{3}{2}C\mp 6)^2 + (2(C\mp 1))^2+C^2=1$,
$\frac{9}{4}(C^2\mp 8C +16)+4(C^2\mp 2C + 1) + C^2=1$,
$\frac{29}{4}C^2\mp 26C +39=0$
$\emptyset$
And the answer there is no such plane. :)
