Suppose that $ ( a,b,c)$ is a solution and $ a\geq b\geq c$. $2=2( 3k+1)^{2} -6( 3k^{2} +2k)\\ =2( a+b+c)^{2} -6( ab+bc+ca)\\ =( a-b)^{2} +( b-c)^{2} +( a-c)^{2}\\ $ We consider two cases (1),(2) : (1) $ a-c=b-c=1$ Since $ a+b+c=3k+1$, we get $ ( a,b,c) =( k+1,k,k)$ (2) $ b-c=1,\ a=b$ Since $ a+b+c=3k+1$, we have $ b+b+( b-1) =3k+1$. However, in this case, we get $ -1\equiv 1\pmod 3$, which is impossible. Conversely, we can verify that $ ( a,b,c) =( k+1,k,k)$ is a solution, of course. Conclusion : $\boxed{( a,b,c) =( k+1,k,k)}$ is the only solution.