数列 $\left\{a_{n}\right\}$ 满足 $3+a_{n}=a_{n+1}(n \in N+)$ 且 $a_{2}+a_{4}+a_{6}=9$, 则 $\log _{\frac{1}{6}}\left(a_{5}+a_{7}+a_{9}\right)=$
$: 3+a_{n}=a_{n+1} \quad \therefore a_{n+1}-a_{n}=3 \quad \therefore$ 数列 $\left\{a_{n}\right\}$ 是以 3 为公差的等差数列, 由等差数列的性质可得, $a_{2}+a_{4}+a_{6}=3 a_{4}=9 \quad \therefore a_{4}=3, a_{7}=a_{4}+3 d=12$ $\therefore \log _{\frac{1}{1}}\left(a_{5}+a_{7}+a_{9}\right)=\log _{\frac{1}{6}} 3 a_{7}=\log _{\frac{1}{6}} 36=-2 \quad$ 故选 $\mathrm{A}$