jmc

Subtracting $c$ from the first equation and $c^2$ from the second, we get $$\begin{array}{rl}a+b& =2-c,\\ a^2+b^2& =12-c^2.\end{array}$$By Cauchy-Schwarz, $$(1+1)(a^2+b^2)=2(a^2+b^2)\ge (a+b{)}^2.$$Substituting for $a+b$ and $a^2+b^2$ gives $$2(12-c^2)\ge (2-c{)}^2,$$which rearranges to $$3c^2-4c-20\le 0.$$This factors as $$(3c-10)(c+2)\le 0,$$so the maximum possible value of $c$ is $\frac{10}{3}$ (which occurs when $a=b=-\frac{2}{3}$) and the minimum possible value of $c$ is $-2$ (which occurs when $a=b=2$). Thus, the answer is $\frac{10}{3}-(-2)=\boxed{\frac{16}{3}}.$