3. Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}=3 \text {. }
$$

Then the range of values for $d$ is $\qquad$ .
3.0 $\leqslant d \leqslant \frac{3}{2}$.

Let $y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \geqslant 0$.
Then $y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \geqslant 0$.
Since the coefficient of $x^{2}$ is $3>0$, we have,
$$
\begin{array}{l}
\Delta=[-2(a+b+c)]^{2}-4 \times 3\left(a^{2}+b^{2}+c^{2}\right) \leqslant 0 \\
\Rightarrow 4(3-d)^{2}-4 \times 3\left(3-d^{2}\right) \leqslant 0 \\
\Rightarrow 2 d^{2}-3 d \leqslant 0 \\
\Rightarrow 0 \leqslant d \leqslant \frac{3}{2} .
\end{array}
$$

When $a=b=c=1$, $d=0$;
When $a=b=c=\frac{1}{2}$, $d=\frac{3}{2}$.