3. Given distinct real numbers $a$, $b$, $c$ satisfying $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=t$.
Find the value of $t$.
By eliminating $b$ and $c$ from the given equation, we get
$$
\left(t^{2}-1\right)\left(a^{2}-t a+1\right)=0 \text {. }
$$

If $a^{2}-t a+1=0$, then
$$
a+\frac{1}{a}=t=a+\frac{1}{b} \Rightarrow a=b \text {, }
$$

which is a contradiction.
Thus, $t^{2}=1 \Rightarrow t= \pm 1$.