3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is

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3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$

we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$.