How to find this point 
Suppose this is you in 3D space, located at position O, looking at a cube, such that A is the closest vertex to O, F is the farthest vertex, and O, A and F are collinear.
Suppose you want to draw segments PX and PY on the cube's faces, such that it looks as a straight line when observed from O.
I guess we can rewrite the last sentence as: P is on the OXY plane.
Since P is on the line segment AB, we can write it as P=A+(B-A).
Hence (if O is (0,0,0)'): (OX x CY)' [A+(B-A)] = 0
how to solve for ?
A, B, X, Y, O are points given as 3D vectors
 A: $$
A+\alpha(B-A)=O+s(X-O)+t(Y-O).
$$
Solve for $\alpha$, $s$, $t$.
If you want a symbolic solution, define $\vec v=(X-O)\times(Y-O)$ and take the scalar product of both sides of the equation with $\vec v$, to obtain:
$$
\alpha(B-A)\cdot\vec v=(O-A)\cdot\vec v.
$$
A: The four points are coplanar, so $$\det {\begin{bmatrix}O & 1 \\ X & 1 \\ Y & 1 \\ P & 1 \end{bmatrix}} = 0.$$ Set $P=(1-\alpha)A + \alpha B$ (an alternate form of your parameterization) and solve for $\alpha$.
A: Let $OP\cap BF=\{G\}$.
Thus, $X$, $Y$ and $G$ they are colinear.
A: 
This is one (not easy) way to build your point $P$ :


*

*let $U$ be the intersection of $(XY)$ with $(FG)$;

*draw the parallel to $(XG)$ going threw $A$ meet $(CD)$ at $W$;

*$(AW)$ is then the intersection of the planes $(ABC)$ and $(OAX)$;

*therefore, lines $(OX)$ and $(AW)$ meet at a point $R$ which is the intersection of line $(OX)$ and plane $(ABC)$, so $R$ is a point of the intersection of planes $(ABC)$ and $(OXY)$;

*this intersection $\Delta$ is the parallel to $(XY)$ going threw $R$;

*$\Delta$ meets $(BC)$ at a point $S$;

*$(SU)$ is then the intersection of planes $(OXY)$ and $(BCF)$;

*as $(BCF)$ and $(AEH)$ are parallel, the intersection of planes $(OXY)$ and $(AEH)$ must be parallel; the point $Y$ being one of this intersection, the parallel to $(US)$ going threw $Y$ is this intersection; it cuts $(AE)$ at the searched for point $P$.


Hope I didn't make a mistake, and that my english is understandable...
