Find location and normal of a plane based on the relative intersection points from 3 lines Here is an illustration of the problem

I have 3 points in a 3d space, in my example the points are: (0, 0, 0), (26, 0, 0) and (13, 27, 0)
I then have a plane with 3 holes in it, with positions relative to the center of the plane, positions of those points are: (-0.5, -0.25, 0), (-0.1, -0.5, 0), (-0.2, 0.5, 0) and a point located 1 unit above the plane (so the relative position of this point to the plane is (0, 0, 1) )
The size of the plane is 1 x 1 if it were layed down on the x/y plane in the 3d space.
How can I find where in 3d space the center point of the plane is located, and the normal of the plane; such that lines can be drawn from all the 3 points through the holes in the plane and intersect at the point above the plane.
If this were in real life I could hold a piece of paper up to my eye and move it around until the 3 holes in the paper line up with the 3 points.
 A: Let the points in the dark grey plane be denoted $(a_x,a_y,a_z),~(b_x,b_y,b_z),~(c_x,c_y,c_z)$. Let the points in the light grey plane be denoted $(d'_x,d'_y,d'_z),~(e'_x,e'_y,e'_z),~(f'_x,f'_y,f'_z)$. Note, however, that the $d',e',f'$ points are given in a different coordinate system corresponding to the small light grey plane, that is rotated and translated away from the coordinate system given by the large dark grey plane. Let's say that the point $\vec{d}\,\!'$ can be denoted by $\vec{d}=(d_x,d_y,d_z)$ in the latter coordinate system (which I will take as the more 'natural' or 'desirable' set of coordinates to work with). Similarly for points $e'$ and $f'$.
Now, suppose that the centre of the small light grey plane is given by $\vec{g}=(g_x,g_y,g_z)$. Then the coordinate transformation from the dark to the light plane is given by multiplying by a rotation matrix, say $\overleftrightarrow{R}$, and then adding a translation by $\vec{g}$. So, we have $\vec{d}\,\!'=\overleftrightarrow{R}\vec{d} +\vec{g}$. What we already know is $d'$ and we want to find $d$, so we note that $d=R^{-1}(d'-g)$. Similarly for $f$ and $g$. This also works to find the 'eye', which in the light grey's coordinate system is $\hat{z}'=(0,0,1)$ and can be converted to the dar grey plane's coordinate system similarly (let's denote it $z$).
Now write down the equation of a line between $a$ and $d$ and demand that $z$ lies on it. Repeat with $b$ and $e$, and with $c$ and $f$. You should get a bunch of equations that you can use to solve for $R$ and $g$, which are what you want. In particular, the normal to the light grey plane should be the third column of $R$.
