1. If positive numbers $a, b$ satisfy $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$, then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$
Given $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k \Rightarrow\left\{\begin{array}{l}4 a=2^{k} \\ 27 b=3^{k} \\ a+b=6^{k}\end{array}\right.$ $\Rightarrow a+b=4 a \cdot 27 b=108 a b \Rightarrow \frac{1}{a}+\frac{1}{b}=108$.