We can see that there is an invariant that if triple $(a,b,c)$ is tranformed to triple $(x,y,z)$ by this operations then pairs $(a,x), (b,y), (c,z)$ have numbers with the same parity, so from triple $(3,5,14)$ we cannot get the triple $(9,8,11)$ because $5$ and $8$ have different parity.

The triples are unordered.

Sorry, I haven't seen that. In such case from that what I've written before we know that there is always exactly one even number in our triple. From that it follows that $2|a^2+b^2-c(a+b)$, so when we change triple $(a,b,c)$ to $(a,b,2a+2b-c)$ then we have invariant that sum of squares is constant modulo $8$. Indeed, $a^2+b^2+(2a+2b-c)^2=4(a^2+b^2-c(a+b))+a^2+b^2+c^2$, and we know that $8$ divides $4(a^2+b^2-c(a+b))$, so this sum of squares doesn't change mod $8$, but $3^2+5^2+14^2$ and $9^2+8^2+11^2$ gives different residues modulo $8$, so there is no possibility to transform first triple to the second one.