3. Given $a, b, c$ are the lengths of the three sides of $\triangle ABC$,
$$
t=\frac{a+b+c}{\sqrt{a b+b c+c a}} \text{.}
$$

Then the range of values for $t$ is
$$
\begin{array}{l}
\text { 3. } \sqrt{3} \leqslant t<2 \text {. } \\
\text { From }(a+b+c)^{2} \geqslant 3(a b+b c+c a) \\
\Rightarrow t=\frac{a+b+c}{\sqrt{a b+b c+c a}} \geqslant \sqrt{3} .
\end{array}
$$

Since $|a-b|<c<a+b$, therefore,
$$
\begin{array}{l}
(a-b)^{2}+c^{2}<2 c^{2}<2 c(a+b), \\
a^{2}+b^{2}+c^{2}<2(a b+b c+c a), \\
(a+b+c)^{2}<4(a b+b c+c a) .
\end{array}
$$

Thus, $t<2$.
Therefore, $\sqrt{3} \leqslant t<2$.