I'm having some trouble in finding the plane which the distance between it and the point $(-1,0,1)$ is: $2$ and also it passes through the points $(0,0,6)$ and $(2,0,3)$.
I tried to start by finding the line that passes through those points $(0,0,6)$ and $(2,0,3)$ so I find this line which the plain consists and I got stuck finding the normal of the plane by the dot product with the direction vector of the line because I don't find a connection to $B$ $(Ax+By+Cz+D=0)$, so therefore $B$ can be anything which because of that I can't find a specific equation for the needed plane or planes for that matter (if because of the distance we will get two options) . Would love some help figuring this one out. thank you :)
Let $ax+by+cz=d$ be the plane in question. It goes thru $(0,0.3)$ and $(2,0,3)\implies a\cdot 0+b\cdot 0+c\cdot 3 = d\implies d = 3c$. Also, $a\cdot 2+ b\cdot 0+ c\cdot 3=d=3c\implies a=0\implies by+cz=3c$ is the equation of the plane. We still need to find $b,c$. The distance is now used: $\dfrac{|b\cdot 0+ c\cdot 1-3c|}{\sqrt{b^2+c^2}}=2\implies 4c^2=4(b^2+c^2)\implies b = 0\implies cz=3c\implies z =3$ is the equation of the plane.
Note: This answer is based on the unedited post. With the edited post, the $(0,0,3)$ is changed to $(0,0,6)$, and the analysis is the same but with a different equation for the plane.