smc

Because $a+b+c$ is odd, $a$, $b$, $c$ are either one odd and two evens or three odds. $\text{Case 1}$: $a$, $b$, $c$ have one odd and two evens. Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even. Hence, $\gcd \left(a,b\right)$ and $\gcd \left(a,c\right)$ are odd, and $\gcd \left(b,c\right)$ is even. Hence, $\gcd \left(a,b\right)+\gcd \left(b,c\right)+\gcd \left(c,a\right)$ is even. This violates the condition given in the problem. Therefore, there is no solution in this case. $\text{Case 2}$: $a$, $b$, $c$ are all odd. In this case, $\gcd \left(a,b\right)$, $\gcd \left(a,c\right)$, $\gcd \left(b,c\right)$ are all odd. Without loss of generality, we assume $$\gcd \left(a,b\right)\le \gcd \left(b,c\right)\le \gcd \left(c,a\right).$$ $\text{Case 2.1}$: $\gcd \left(a,b\right)=1$, $\gcd \left(b,c\right)=1$, $\gcd \left(c,a\right)=7$. The only solution is $(a,b,c)=(7,9,7)$. Hence, $a^2+b^2+c^2=179$. $\text{Case 2.2}$: $\gcd \left(a,b\right)=1$, $\gcd \left(b,c\right)=3$, $\gcd \left(c,a\right)=5$. The only solution is $(a,b,c)=(5,3,15)$. Hence, $a^2+b^2+c^2=259$. $\text{Case 2.3}$: $\gcd \left(a,b\right)=3$, $\gcd \left(b,c\right)=3$, $\gcd \left(c,a\right)=3$. There is no solution in this case. Therefore, putting all cases together, the answer is $179+259=\boxed{438}$.