In general, the answer to your problem is no, it is not possible to do so. There are however variants that are possible if we know more about the transformation.
Both I get into detail, some notation and two observations:
Notation: We know 3 points, $a$,$b$,$c$ and a projection $P$ onto a 2d plane through the origin (so $P$ is a matrix with $P^2=P$ and $P$ is of rank 2). There is an unknown matrix $M$, but we know the three points $PMa, PMb$ an $PMc$.
I assume your original points $a$,$b$,$c$ are linearly independent, otherwise (for example if one of them is (0,0,0)) it will clearly not work.
It does not matter if we recover the 3d image points $Ma,Mb$ and $Mc$ or the transformation matrix. If we have the matrix we can apply it to $a,b,c$ in order to get the points. If we have the points, we can recover the matrix, since $a,b,c$ form a basis and any matrix is uniquely given by its image in any basis (see footnote 1).
Now to the actual problem. By a simple heuristic, we know three 2d coordinates, that is 6 independent values (the initial points $a,b,c$ do not count, see footnote 2), but we want to recover all 9 independent entries of $M$. There is obviously a mismatch here. This is no proof, but it tells us to search for a counterexample:
The problem essentialy is that we do not know the height the points $Ma,Mb$ and $Mc$ have above the plane. In fact let $v\neq 0$ be in the kernel of $P$, that is $Pv = 0$ (such a $v$ exists, since $P$ is of rank 2). Let $A$ be a matrix whose collumns are all multiples of $v$. Then the image of $A$ is the kernel of $P$ and we have
$$ P(M+A) x = PMx+PAx = PMx$$
for any $x$. So using only the data given we cannot distinguish $M$ and $M+A$ as transformations, thus the answer is no, it is not possible.
As I said in the beginning, however there variants, for which this is possible. For example, if we know that $M$ is a rotation, we search for fewer independent values, so it should be possible. As said in the footnotes, we know the matrix $PM$. Along the axis of rotation we know that $Mx=x$ and thus $PMx=Px$, which is a linear equation with a 1d space of solutions (except when $M$ is the identity, but that is exactly the case if $PM=P$, so we can exclude this) which then give us this axis.
Let us call the axis $v$. Now we have two cases:
The axis of rotation does not lie in the plane on which we project. Then $P$ restricted to vectors from the plane of rotation $v^\perp$ is invertible, so we can recover how $M$ transforms the plane $v^\perp$ and thus recover $M$.
The axis of rotation lies in the plane on which we project. Then there is a vector $x$ in $v^\perp$ such that $PMx=0$, since $x$ is rotated to a vector perpendicular to the plane. But then we can find $x$ and recover the angle of rotation this way, also finding $M$.
1.More detail, by linear independence, there are $r,s,t$ such that $(1,0,0)^T=r a+sb+tc$, but then $M(1,0,0)^T = rM a+sMb+tMc$ so we recover the first collumn of $M$. Then continue in the same way for $(0,1,0)$ and $(0,0,1)$.
- The points $a,b,c$ do not contain any information about $M$, they only serve to determine the images. However note that using the same idea as in footnote 1, we can actually recover the matrix $PM$ from the initial information. We cannot get $M$ from this, since $P$ is of rank 2 and thus not invertible.