3. If $a, b, c, x, y, z$ are real numbers, and $a^{2}+b^{2}+c^{2}=25, x^{2}+y^{2}+z^{2}=36, a x+b y+c z=$ 30, then $\frac{a+b+c}{x+y+z}=$ $\qquad$
3. $\frac{5}{6}$.

From the given conditions, we have
$$
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
=(a x+b y+c z)^{2} .
\end{array}
$$

By the condition for equality in the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\frac{a}{x}=\frac{b}{y}=\frac{c}{z} . \\
\text { Let } \frac{a}{x}=k, \text { then } \frac{a+b+c}{x+y+z}=k .
\end{array}
$$

Substituting $a=kx, b=ky, c=kz$ into $a x+b y+c z=30$, we get
$$
\begin{array}{l}
k\left(x^{2}+y^{2}+z^{2}\right)=30 . \\
\therefore k=\frac{30}{36}=\frac{5}{6}, \text { i.e., } \frac{a+b+c}{x+y+z}=\frac{5}{6} .
\end{array}
$$