I have three planes in general position, and in each plane an arbitrary point is selected : this gives us three points $R,S,T$. Is it possible to construct the intersection lines of the $(RST)$ plane with each of the original planes using only a straightedge (so, we can only draw parallels and locate intersection points of two secant lines)?
One could formalize the problem as follows : we have a set $\cal S$ of lines and points, defined as the smallest set containing $R,S,T$ and the three intersection lines of the original planes, and closed with respect to line intersection (if two secant lines $D,D' \in {\cal S}$, then the intersection point $D\cap D'$ is in $\cal S$) and parallel straight lines (if $D\in S$ and $p\in S$ then the parallel to $D$ passing by $p$ is in $\cal S$). The question can then be restated as, does $\cal S$ contain the intersection lines of $(RST)$ with the original planes.
The answer to this question is YES (and I got it from a MO user in a comment shortly after having copied this question to MO ; the MO version is now deleted as it is not research-level).
Let us denote the $(RST)$ plane by $\Pi$. Let us perturbate the initial problem slightly, replacing $R$ with a point $R'$ on one of the lines defining the "yellow" plane containing $R$.
Then the intersection of $\Pi'=(R'ST)$ with our initial three "coloured" planes is easy to construct, as shown below :
Next, notice that the intersection point of $(ST)$ and $(R'H)$ (let us call it $D$) is remarkable in that he is on $\Pi$, $\Pi'$ and the "yellow" plane all at once :
It follows that the intersection of $\Pi$ with the yellow plane is simply $(DR)$. The other intersections are constructed similarly.