I have 3 simple straight lines, 2 are nearly horizontal and parallel and 1 is nearly vertical (blue in the diagram).
I have a triangle (orange) which is fixed in shape by the lengths of its 3 sides.
There will be only one postion of the triangle where the 3 verticies are directly on the 3 lines. How is this calculated?
The calculation needs to be done frequently for slightly varying line lengths of the triangle.
The lengths of the blue lines are not important but typically the variations expected would fit in the short distances shown.
It is only the 3 contacts condition that is required, the triangle is not allowed to rotate / move way from this.
I would like to know the 3 contact points on the 3 blue lines.
Not a solution (yet).
This is a hard problem with strong hints of mechanism synthesis and rigid body kinematics.
Ideal Solution
I think what you are looking for is this
Three fixed lines form a triangle ABC in space. Let's call these the walls. A smaller rigid body triangle PQR is inscribed inside the walls. The sides if the inscribed triangle are known as $\ell_{PQ}$, $\ell_{QR}$ and $\ell_{RP}$.
Statics
In this ideal solution, each contact point P, Q and R has a normal vector (red dashed arrows below) describing the direction of where contact forces could develop.
Each arrow is perpendicular to the wall in contact and goes through the respective contact point.
Where any two these arrows meet, for example at GPQ above, it designates the instant center of rotation for the side of the triangle connecting the two contact points (side PQ here).
This means that points P and Q both are going to slide along the wall when the rigid body is rotated about GPQ.
This is a kinematic condition to maintain contact with the walls at any time
We exploit this geometry to nudge the body in order to make all three points contact the walls
Kinematics
Now imagine a situation where point Q is not quite in contact yet
We can rotate the about the point GRP where the side RP is going to slide along their walls in order to bring Q closer to its wall. Draw radiuses from GRP to P, Q and R and show the direction of motion for each triangle corner in blue arrows above.
Now if you rotate by a finite angle $\Delta \theta > 0$ the points are going to follow little arcs, and not lines. This means that yes point Q is going to get closer to the wall, but points P and R are actually going to come off the wall by a tiny amount.
This means that in practice you will have to go around each point, doing rotation such as that point meets the wall. Then rinse and repeat with each point on the triangle.
At some point, you will be close enough to the wall to have a satisfactory answer, but you will never get the exact position where all three points are going to be on each wall.
Please note that if all three contact normals come together through the same point, then the body is not stable as any torque applied would twist the part and it would just happily slide along the walls.