3. Given that $a, \dot{c}, c$ are not all equal real numbers. Then, the equation about $x$: $x^{2}+(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right)=0(\quad)$.
(A) has two negative roots
(B) has two positive roots
(C) has two real roots with the same sign
(D) has no real roots.
3. (D).
$$
\begin{aligned}
\Delta= & (a+b+c)^{2}-4\left(a^{2}+b^{2}+c^{2}\right) \\
= & a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
& -4\left(a^{2}+b^{2}+c^{2}\right) \\
= & -(a-b)^{2}-(b-c)^{2}-(c-a)^{2} \\
& -\left(a^{2}+b^{2}+c^{2}\right) \\
& <0,
\end{aligned}
$$
$\therefore$ the original equation has no real roots.