$2$-dimensional to $3$-dimensional projection I have the coordinates of three vertices of a triangle on a $2$-dimensional image plane - $(x_1,y_1), (x_2,y_2)$ and $(x_3,y_3)$. Their $3$-dimensional real world projection points are also known - $(x_1',y_1',z_1'), (x_2',y_2',z_2')$ and $(x_3',y_3',z_3')$ respectively.
Now I have to find the $3$-dimensional projection of another arbitrary point $(x_c, y_c)$, how can I find them, if it is a linear projection? 
Note: I do not have the information about $d$, where $d$ is distance between COP(center of projection) and PP(Projection plane/monitor). Then I assume that, I could use perspective projection equation.
 A: Where is the two-dimensional plane in the three-dimensional space and how do the coordinates of the plane and space relate to each other?
In general, is the point $(x,y)$ in the plane simply at $(x,y,0)$ in space?
Once you have determined how to find the full spatial coordinates of each point in the plane (e.g., by setting $z= 0,$ or whatever is necessary),
find the three-dimensional coordinates of the three points of the original triangle,
$A,$ $B,$ and $C.$
If these are being projected to $A',$ $B',$ and $C',$ respectively,
then the lines $AA',$ $BB',$ and $CC'$ all meet at a single point.
Find that point.
If the three lines do not meet then perhaps the projection is not from $A$ to $A'$ but from $A$ to $B'$ instead. There are six ways the points could have been permuted; if none of them results in three concurrent lines of projection then the "projection" was not as it was supposed to be after all.
A: Say the linear transformation is$$(x,y)\mapsto(ax+by,cx+dy,ex+fy)$$You can solve for the $6$ unknowns $a,b,c,d,e,f$ using any two points that give you $6$ independent equations. Say you chose $(x_1,y_1),(x_2,y_2)$, then$$\begin{matrix}ax_1+by_1=x_1'&ax_2+by_2=x_2'\\cx_1+dy_1=y_1'&cx_2+dy_2=y_2'\\ex_1+fy_1=z_1'&ex_2+fy_2=z_2'\end{matrix}$$Once you find $a,b,c,d,e,f$, you can transform any point in $\Bbb R^2$. 
