jmc

Let $m=121^2+233^2+345^2$ and $n=120^2+232^2+346^2$. By the Euclidean Algorithm, and using the difference of squares factorization, $$\begin{array}{rl}\text{gcd}(m,n)& =\text{gcd}(m-n,n)\\ & =\text{gcd}(n,121^2-120^2+233^2-232^2+345^2-346^2)\\ & =\text{gcd}(n,(121-120)(121+120)\\ & +(233-232)(233+232)\\ & -(346-345)(346+345))\\ & =\text{gcd}(n,241+465-691)\\ & =\text{gcd}(n,15)\end{array}$$We notice that $120^2$ has a units digit of $0$, $232^2$ has a units digit of $4$, and $346^2$ has a units digit of $6$, so that $n$ has the units digit of $0+4+6$, namely $0$. It follows that $n$ is divisible by $5$. However, $n$ is not divisible by $3$: any perfect square not divisible by $3$ leaves a remainder of $1$ upon division by $3$, as $(3k\pm 1{)}^2=3(3k^2+2k)+1$. Since $120$ is divisible by $3$ while $232$ and $346$ are not, it follows that $n$ leaves a remainder of $0+1+1=2$ upon division by $3$. Thus, the answer is $\boxed{5}$.