17. Let $a, b, c \in \mathbf{R}$, and
$$
a+b+c=2, a^{2}+b^{2}+c^{2}=12 \text {. }
$$

Then the difference between the maximum and minimum values of $c$ is ( ).
(A) 2
(B) $\frac{10}{3}$
(C) 4
(D) $\frac{16}{3}$
(E) $\frac{20}{3}$
17. D.

From the problem, we have $\left\{\begin{array}{l}a+b=2-c, \\ a^{2}+b^{2}=12-c^{2} .\end{array}\right.$ Then,
$$
\begin{array}{l}
a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=\frac{(2-c)^{2}-\left(12-c^{2}\right)}{2} \\
=c^{2}-2 c-4 .
\end{array}
$$

Thus, the quadratic equation with roots $a, b$ is
$$
x^{2}+(c-2) x+\left(c^{2}-2 c-4\right)=0 \text {. }
$$

Its discriminant is
$$
\begin{array}{l}
\Delta=(c-2)^{2}-4\left(c^{2}-2 c-4\right) \geqslant 0 \\
\Rightarrow 3 c^{2}-4 c-20 \leqslant 0 \Rightarrow-2 \leqslant c \leqslant \frac{10}{3} .
\end{array}
$$

Therefore, the difference between the maximum and minimum values of $c$ is $\frac{10}{3}+2=\frac{16}{3}$.