I'm trying to get the width of a rotated plane, but my knowledge of trig functions didn't really help me get what I want.
I have a plane, that is $310$ units wide, and is $200$ units away from the camera. If I rotate the plane around the $Z$ axis, it becomes narrower in the camera's perspective. What I want to calculate is, how much I should rotate the plane, for it be $100$ units wide in the camera's perspective. The origin if rotation is the plane's left edge.
On
If I understand your set-up correctly (and I am not sure I do), then you need to rotate the plane back away from the camera by an angle of $85.7$ degrees from its straight-on position.
It would be quicker for you to check this with your camera than for me to enter the solution, and of course I may have misunderstood, so perhaps you would like to try it and let me know if it works.
Here is a diagram, from Geogebra, that shows an answer. Your question is still ambiguous on the exact placement of the "plane", so I made the initial placement symmetric with the camera. All linear units here are reduced by a factor of $100$, which will not change the final angle. The view of this diagram is vertically above the camera and "plane" so the "plane" appears to be a horizontal blue line segment $\overline{AB}$. The left edge of the "plane" is at the origin.
You can see that is a large angle, over $87°$, and you must rotate back. This is because the "plane" is so close to the camera, compared to its width.
This can be solved exactly, but using analytic geometry rather than trigonometry. The line $\overleftrightarrow{CF}$ has the equation
$$y=-\frac{40}{11}(x-1)$$
and the circle has the equation
$$x^2+y^2=\left(\frac{31}{10}\right)^2$$
This is best solved numerically, but if you really want part of the exact answer it is
$$x=\frac{1600}{1721}-\frac{11}{17210}\sqrt{1493881}$$
From that you can find $x$, then the desired angle $\tan^{-1}\frac yx$. The final answer is about $87.2547°$.
Does this meet your needs?