5. Given real numbers $a>b>c$, and satisfying:
$$
a+b+c=1, a^{2}+b^{2}+c^{2}=3 \text {. }
$$

Then the range of $s=b+c$ is ( ).
(A) $-\frac{2}{3}<s<2$
(B) $-\frac{2}{3}<s<1$
(C) $-\frac{2}{3}<s<\frac{2}{3}$
(D) $-\frac{2}{3}<s<0$
5. D.

Find the range of $a$.
From the problem, we know
$$
s=b+c=1-a, \quad b c=a^{2}-a-1 \text {. }
$$

Then the quadratic equation with $b, c$ as real roots is
$$
t^{2}+(a-1) t+\left(a^{2}-a-1\right)=0 \text {, }
$$

This equation has two distinct real roots less than $a$, thus,
$$
\left\{\begin{array}{l}
\Delta=(a-1)^{2}-4\left(a^{2}-a-1\right)>0, \\
(a-b)(a-c)>0, \\
(a-b)+(a-c)>0 .
\end{array}\right.
$$

Solving this, we get $1<a<\frac{5}{3}$.
Furthermore, when $1<a<\frac{5}{3}$, all the above inequalities hold, and there exist real numbers $a, b, c$ that satisfy the problem.