3. A certain temple has three bronze bells, A, B, and C. Bell A rings once every $4 \mathrm{~s}$, Bell B rings once every $5 \mathrm{~s}$, and Bell C rings once every $6 \mathrm{~s}$. At the arrival of the New Year, the three bells ring simultaneously and stop ringing at the same time. A person hears a total of 365 bell rings. If during this period, Bell A, Bell B, and Bell C ring $x$ times, $y$ times, and $z$ times respectively, then $x+y+z=$ $\qquad$
3.484 .

Let the duration of the bell ringing be $60 k \mathrm{~s}$, then during this period, the number of times the first clock rings is $x=\frac{60 k}{4}+1=15 k+1$; the number of times the second clock rings is $y=\frac{60 k}{5}+1=12 k+1$; the number of times the third clock rings is $z=\frac{60 k}{6}+1=10 k+1$; the number of times the first and second clocks ring simultaneously is $\frac{60 k}{20}+1=3 k+1$; the number of times the first and third clocks ring simultaneously is $\frac{60 k}{12}+1=5 k+1$; the number of times the second and third clocks ring simultaneously is $\frac{60 k}{30}+1=2 k+1$; the number of times the first, second, and third clocks ring simultaneously is $\frac{60 k}{60}+1=k+1$. According to the problem, we have
$$
\begin{aligned}
365= & (15 k+1)+(12 k+1)+(10 k+1)- \\
& (3 k+1)-(5 k+1)-(2 k+1)+(k+1) \\
= & 28 k+1 .
\end{aligned}
$$

Therefore, $k=13$.
Substituting $k$ into the expressions for $x, y, z$ we get
$$
\begin{array}{l}
x=15 k+1=196, y=12 k+1=157, \\
z=10 k+1=131 .
\end{array}
$$

Thus, $x+y+z=484$.