WLOG, let's assume the ball is initially in cup 1 on the left. Then, the probability of the final position of the ball is:
$P(L)=30\%$, $P(M)=35\%$, $P(R)=35\%$.
There are numerous ways of counting this, but here is a neat method using a triangle!
But before that
let's calculate the number of switching of each pair of cups based on the given conditions.
Let $A$ denote the action of switching the left and middle cups, $B$ the switching of middle and right cups, and $C$ the switching of the left and middle cups. Assume $x$$A$'s, $y$$C$'s and $z$$B$'s are performed. Then, we have $$x+y=5, x+z=4, y+z=3,$$which produces$$x=3, y=2, z=1.$$
Now let's introduce the triangle!
In the picture, the black-and-white triangle with 1 2 3 denotes the cups, and the red, green, and blue circles denote the positions. Consider every move as an action performed on the triangle, such as rotating or flipping it, and ensuring that its three vertices land in the three circles.
For example, consider the action switching the middle and right cups. That action can be represented using our triangle in the following way:
With the help of a triangular slip of paper, it can be shown that all "switchings" is equivalent to flipping the triangle, and performing two "switchings" is equvalent to a rotation by a certain degree, namely: $$AA=BB=\text{0 degrees rotation}$$$$AB=BC=CA=\text{120 degrees, clockwise rotation}$$$$BA=CB=AC=\text{120 degrees, counterclockwise rotation}$$.
Note: the notation $XY$ where $X$ and $Y$ are both actions represents the resultant action of first performing action $X$ then performing action $Y$.
Now what?
Notice that we can perform 3$A$'s, 2$C$'s and 1$B$ is total-that's 6 "switchings", equivalent to 3 rotations. Hence, by splitting the string of 6 "switchings" into 3 groups of 2 "switchings", or 3 rotations, the problem immediately becomes much more approachable!
And here's the final part of the analysis.
Here's the string of actions: (xx) (xx) (xx). Consider overall action performed on the triangle in the following 4 cases:
Case 1: (AA) and (CC) are in the string of actions.
Then the remaining (xx) must be (A&B). (The notation (X&Y) means either (XY) or (YX).) There are 6 ways to arrange the 3 groups of actions no matter what the remaining (xx) is. Also, the final outcome is completely dependent on the outcome of the remaining (xx), as actions (AA) and (CC) do nothing.
Conclusion: 6 unique strings of actions with the resultant action being 120 degrees, clockwise rotation; 6 unique strings of actions with the resultant action being 120 degrees, counterclockwise rotation.
Case 2: (AA), but not (CC), is in the string of actions.
Then the remaining 2 (xx)'s must be (A&C) and (B&C). There are 6 ways to arrange the 3 groups of actions no matter what the 2 remaining (xx) is. Also, the final outcome is completely dependent on the outcome of the 2 remaining (xx)'s, as action (AA) does nothing. The resultant action of the two (xx)'s can either be 120 degrees, clockwise rotation (when both actions rotate the triangle 120 degrees counterclockwise), 120 degrees, counterclockwise rotation (when both actions rotate the triangle 120 degrees clockwise), or nothing (when the two actions are in opposite directions, hence cancelling out with each other).
Conclusion: 6 unique strings of actions with the resultant action being 120 degrees, clockwise rotation; 6 unique strings of actions with the resultant action being 120 degrees, counterclockwise rotation; 12 unique strings of actions with the resultant action being doing nothing.
Case 3: (CC), but not (AA), is in the string of actions.
This is impossible by using the pigeonhole principle on action $A$.
Case 4: Neither (CC) nor (AA) is in the string of actions.
In this case, the three (xx)'s must each contain an $A$, with the other element being $B$ in two (xx)'s and $C$ in one (xx). In this case, there are three arrangements of the (xx)'s: (A&B) (A&C) (A&C), (A&C) (A&C) (A&B), (A&C) (A&B) (A&C). For each arrangement, there are 8 choices of the order of X and Y in (X&Y), of which two rotates the triangle a full 360 degrees back to its original position (all clockwise/counterclockwise rotations), 3 rotates the triangle 120 degrees clockwise (2 clockwise, 1 counterclockwise rotations), and 3 rotates the triangle 120 degrees counterclockwise (1 clockwise, 2 counterclockwise rotations).
Conclusion: 9 unique strings of actions with the resultant action being 120 degrees, clockwise rotation; 9 unique strings of actions with the resultant action being 120 degrees, counterclockwise rotation; 6 unique strings of actions with the resultant action being doing nothing.
And finally, in conclusion to the casework above:
There are 21 unique strings of actions with the resultant action being 120 degrees, clockwise rotation; 21 unique strings of actions with the resultant action being 120 degrees, counterclockwise rotation; 18 unique strings of actions with the resultant action being doing nothing.
This can be verified to be correct, as $$21+21+18=60=\dfrac{6!}{3!\cdot 2!},$$which is the total number of distinct strings of actions.
Conclusion:
In all the possible strings of actions, $18/60=30\%$ doesn't do anything, leaving cup 1 (which we assumed to contain the ball) in its original position on the left. $21/60=35\%$ rotates the triangle clockwise by 120 degrees, meaning that cup 1 is now on the right. $21/60=35\%$ rotates the triangle counterclockwise by 120 degrees, meaning that cup 1 is now on the left.