If s1%3=s2%3=s3%3=x, then (s1+s2+s3)%3=(3*x)%3=0.

If they are all different, then {s1,s2,s3}%3={1,2,3}, then (s1+s2+s3)%3=0

This means that if you can cut the array as the problem requieres, then the sum of s1+s2+s3 is 0 mod3, an easy check you can see that if two of s1,s2,s3 are the same mod3 while the other is not, then s1+s2+s3 is never 0 mod3.

This means that you can cut the array if and only if (s1+s2+s3)%3=0, which is the sum of all aimod3. Now, your solution says that if you can cut the array, then the solution is 1,n-1. This means that you take 1 element for prefix and suffix, and the rest for the middle. Clearly, if a1=an-1=x mod 3, then the sum of the middle is also x mod3 as the total sum is 0 mod 3, also if a1!=an-1 mod 3, it leads to the same conclusion.

I think I didn't get silly in my proof, and I think it leads to this conclusion: If you are able to cut the array, i.e the sum of all ai mod3=0, then you can cut in any way