It started with this video: https://youtu.be/aLaqMreVj9o?t=56
showing a "Virtual Rolling Contact Joint" - VRCJ:
and then pmesten18 made a model in Fusion 360 here: https://forums.autodesk.com/t5/fusion-360-design-validate/how-would-you-create-the-functional-joints-in-this-spherical/m-p/9072192#M198903
The model it not working. It is unknown if it is because the VRCJ is mathematical impossible!? -or if the software locks when trying to solve the position. Obviously the VRCJ shown in the video is working, but that could be because of slack in the pivot joints.
If the three arms are able to twist, then Fusion 360 can model the VRCJ just fine. It is shown also here: https://forums.autodesk.com/t5/fusion-360-design-validate/how-would-you-create-the-functional-joints-in-this-spherical/m-p/9072192#M198903
Is the ideal mathematical VRCJ possible? Does it need special angles/lengths of the arms?
Edit:
After the answer from @Vasily Mitch I modeled one arm, of the VRCJ, and a rod connecting the centers of the two discs. The rod is connected with a ball joint in each end, and is only there to enforce a fixed distance between the two discs (a needed condition for a VRCJ).

The VRCJ can swing to the side in one projection(dimension):

and to the side in another:


Let's consider one general element of the mechanism.
We have two principal motions here. In one projection there is no doubt that the motion can be depicted as virtual ball rolling. Because two centers of the ball are connected with the constant length.
In other projection, things are not obvious.
So we reformulate this question in a simpler problem: is there 2 points inside 2 balls, so when the balls roll, the distance between the points remain the same?
Let coordinate of this point in ball's frame of reference be $(x,y)$ and two balls of radius $1$ rolled on the angle $\varphi$. Then if point of contact is the origin and $Ox$ goes through centers, coordinates of the points are: $$ P_L=(-1+x\sin\varphi+y\cos\varphi,-x\cos\varphi+y\sin\varphi),\\ P_R=( 1+x\sin\varphi-y\cos\varphi, x\cos\varphi+y\sin\varphi) $$
Square distance between the points are: $$ \frac14 d^2 = \frac14|P_L-P_R|^2 = x^2\cos^2\varphi + (1-y\cos\varphi)^2 = 1 - 2y\cos\varphi + (x^2+y^2)\cos^2\varphi. $$
Unfortunately, this expression doesn't depend on $\varphi$ only when $x=y=0$ (case of the first projection).
So I believe two things are happening on the video. First, there is likely a small slack in pivots allowing for the motion. Second, the motion itself is not mathematically a virtual roll (for, example distance between balls is varying a bit).