Take a look at your last spring composed of 3 vertices. There are two end vertices, and one in the middle. If we lay the vertices all out on a line there is no bend. Then create another spring between the first and last vertex (so it passes through the middle one). This new spring is sometimes called the "bend spring", or something like that. Whenever the 3-vertex chain bends this new spring will push the bend to straighten out.
Yes... except that for small bending angles, the displacement of the end particles (up/down in the OP's diagram) will be mostly orthogonal to the direction along which the spring acts (left-right). So I'd expect the proposed new spring would be effective at preventing substantial bends, but not very effective for keeping a plane flat when it is supposed to not bend very much at all.
If you want to represent very stiff planes, then it might be better to consider adding a spring between the centre particle and the mid-point between the end particles. You will then need to consider how to map the force applied to the mid point to the particles themselves (trivial if all the links have the same rest length, and the particles have the same mass) - you need to make it so that momentum is preserved. If you do this then for small angles, the spring force will be approximately proportional to the bending angle, which is probably what is desired.
Alternatively, it might be even better and just as quick to do the job properly: labelling the particles 1-2-3, take the cross-product between the directions 1-2 and 2-3. This is proportional to the torque vector you want to apply, and the pairs of linear forces will be given by the cross-product between this and each of the deltas 1-2, and 2-3 (divided by the squared delta magnitudes) - since torque = delta x force. You apply a force f1 to particle 1, and apply -f1 to particle 2. Similarly apply f3 to particle 3 and -f3 to particle 2. This preserves momentum. You don't need any trig functions for this, and if you can precalculate the delta lengths, then you might not need any square roots either.