3. Given that $a$ is a constant, and real numbers $x, y, z$ satisfy
$$
(x-1)^{2}+(y-\sqrt{5})^{2}+(z+1)^{2}=a
$$

when, $-8 \leqslant 4 x-\sqrt{5} y+2 z \leqslant 2$. Then $a=$ $\qquad$
3. 1.

Let $4 x-\sqrt{5} y+2 z=k$. Then $-8 \leqslant k \leqslant 2$.
From the given equation, we have
$$
\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}=1 \text {. }
$$

Using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}\right] \\
\geqslant(4 x-4-\sqrt{5} y+5+2 z+2)^{2} \\
\Rightarrow 25 a \geqslant(k+3)^{2} \\
\Rightarrow|k+3| \leqslant 5 \sqrt{a} .
\end{array}
$$

Since $-8 \leqslant k \leqslant 2$, we have
$$
-5 \leqslant k+3 \leqslant 5 \Rightarrow|k+3| \leqslant 5 \text {. }
$$

Therefore, $5 \sqrt{a}=5 \Rightarrow a=1$.