Numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=4$, then the maximum value of $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ is
12. 12.

Garden $(a \quad b)^{\prime}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-(2 a b+2 b c+2 c a)=3\left(a^{2}+b^{2}+c^{2}\right)$ $-(a-b+c)^{2} \leqslant 3 \times 4=12$. When $a+b+c=0$ and $a^{2}+b^{2}-c^{2}=4$, equality is achieved (for example, take $a=0$, $b=-(-\sqrt{2})$.