Uniquely determining the three points which satisfy the constraints Given two fixed lines and three points $P,Q,R$ on one line . Goal is to uniquely find three points on the other line such that they are at equal distances. ($ST=TU$) and that the line joining those points to the intial points intersect at one point (i.e. $PS,QT,RU$).Progress: idea needed  []
 A: 
We give below a construction, which consists of two parts, to determine the locations of the three points $S$, $T$, and $U$ on the given line $GH$. In the first part, we construct the point $M$ on the given line $EF$ such that,
$$\dfrac{MP}{PQ} = \dfrac{MR}{RQ}.$$
What we have obtained here is four collinear points $M$,  $P$, $Q$, and $R$ in a harmonic range. In the second part, we use the constructed point $M$ to locate the three points $S$, $T$, and $U$ on $GH$.
$\underline{\text{Part 1}}$
We take an arbitrary point $A$ as shown in the in $\mathrm{Fig.\space 1}$. After joining $A$ to $Q$, a point $B$ is marked on $AQ$. Then, we draw the lines $AP$, $AR$, $RB$, and $PB$. The two pairs of lines {$AP$, $RB$} and {$AR$, $PB$} intersect with each other at $D$ and $C$ respectively. Finally, we join $C$ to $D$ and extend it to meet $EF$ at $M$.
$\underline{\text{Part 2}}$
After constructing a line parallel to $GH$ through the point $M$, we select an arbitrary point $O$ on it. To complete the construction, we draw the three lines $PO$, $QO$, and $RO$ and extend them to cut $GH$ at $S$, $T$, and $U$. The three points satisfy the constrain,
$$ST = TU.$$
$\underline{Note\space 1}:\space$ As you have already noticed, the point $O$ can be chosen anywhere on the line that goes through $M$. For each $O$, a unique point triplet {$S$, $T$, $U$} obeying the constrain $ST = TU = k$ can be obtained. However, the value of $k$ differs from triplet to triplet.
$\underline{Note\space 2}:\space$ As shown in $\mathrm{Fig.\space 2}$, you may sometimes find that $M$ lies beyond the point $N$, which is the point of intersection between the given lines $EF$ and $GH$. In such instances, the point $O$ does not lie in the interior of $EF$ and $GH$. Nevertheless, the three points $S$, $T$, and $U$ satisfy the constrain $ST = TU.$
$\underline{Note\space 3}:\space$ The way OP used the word $\pmb{uniquely}$ in the problem statement gives the wrong impression that there exists only one $O$ and one set of point triplet {$S$, $T$, $U$} that satisfy the given constrain. On the contrary, there are infinite number of such sets of points.
