3. Non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=10$, find the maximum and minimum values of $\sqrt{6-x^{2}}+\sqrt{6-y^{2}}+\sqrt{6-z^{2}}$.
By the Cauchy-Schwarz inequality, we have
$$
\left(\sqrt{6-x^{2}}+\sqrt{6-y^{2}}+\sqrt{6-z^{2}}\right)^{2} \leqslant 3\left[\left(6-x^{2}\right)+\left(6-y^{2}\right)+\left(6-z^{2}\right)\right]
$$
$=3\left[18-\left(x^{2}+y^{2}+z^{2}\right)\right]=24$, equality holds when $x^{2}=y^{2}=z^{2}=\frac{10}{3}$.
Therefore, the maximum value of $\sqrt{6-x^{2}}+\sqrt{6-y^{2}}+\sqrt{6-z^{2}}$ is $2 \sqrt{6}$.
On the other hand, $\sqrt{a}+\sqrt{b} \geqslant \sqrt{a+b}$, equality holds when $ab=0$. Without loss of generality, assume $x \leqslant y \leqslant z$, then $0 \leqslant x^{2} \leqslant \frac{10}{3}$, thus we have
$$
\begin{array}{l}
\sqrt{6-x^{2}}+\sqrt{6-y^{2}}+\sqrt{6-z^{2}} \geqslant \sqrt{6-x^{2}}+\sqrt{12-\left(y^{2}+z^{2}\right)} \\
=\sqrt{6-x^{2}}+\sqrt{12-\left(x^{2}+y^{2}+z^{2}\right)+x^{2}}=\sqrt{6-x^{2}}+\sqrt{2+x^{2}}
\end{array}
$$

And $\left(\sqrt{6-x^{2}}+\sqrt{2+x^{2}}\right)^{2}=8+2 \sqrt{\left(6-x^{2}\right)\left(2+x^{2}\right)}=8+\sqrt{-x^{4}+4 x^{2}+12}$ $=8+\sqrt{-\left(x^{2}-2\right)^{2}+16}$, from the graph of the quadratic function, we know that when $x=0$, $\sqrt{6-x^{2}}+\sqrt{2+x^{2}}$ reaches its minimum value of $\sqrt{6}+\sqrt{2}$.
In summary, the maximum value of $\sqrt{6-x^{2}}+\sqrt{6-y^{2}}+\sqrt{6-z^{2}}$ is $2 \sqrt{6}$, and the minimum value is $\sqrt{6}+\sqrt{2}$.