11. According to the problem, we have $a_{n+1}=-\frac{1}{2} a_{n}+\frac{1}{3^{n}}$.

Thus, $3^{n+1} a_{n+1}=-\frac{1}{2} \times 3^{n+1} a_{n}+3$.
Let $b_{n}=3^{n} a_{n}$.
Then $b_{n+1}=-\frac{3}{2} b_{n}+3$
$$
\begin{aligned}
\Rightarrow & b_{n}-\frac{6}{5}=-\frac{3}{2}\left(b_{n-1}-\frac{6}{5}\right) \\
& =\left(b_{1}-\frac{6}{5}\right)\left(-\frac{3}{2}\right)^{n-1} \\
& =\left(3 a_{1}-\frac{6}{5}\right)\left(-\frac{3}{2}\right)^{n-1} \\
\Rightarrow & a_{n}=\frac{b_{n}}{3^{n}}=\frac{1}{3^{n}}\left(\frac{6}{5}+\left(3 a_{1}-\frac{6}{5}\right)\left(-\frac{3}{2}\right)^{n-1}\right) \\
& =\left(-\frac{1}{2}\right)^{n-1}\left(\frac{2}{5}\left(-\frac{2}{3}\right)^{n-1}+\left(a_{1}-\frac{2}{5}\right)\right) .
\end{aligned}
$$

If $a_{1}-\frac{2}{5} \neq 0$, then when $n$ is sufficiently large,
$$
\begin{array}{l}
\left|\frac{2}{5}\left(-\frac{2}{3}\right)^{n-1}\right|<\left|a_{1}-\frac{2}{5}\right| . \\
\text { Also, } \frac{2}{5}\left(-\frac{2}{3}\right)^{n-1}+\left(a_{1}-\frac{2}{5}\right) \text { has the same sign as } a_{1}-\frac{2}{5} \text {, }
\end{array}
$$

Therefore, $\left\{a_{n}\right\}$ alternates in sign and is not a monotonic sequence.
When $a_{1}=\frac{2}{5}$, $a_{n}=\frac{2}{5}\left(\frac{1}{3}\right)^{n-1}$ is a decreasing sequence.
11. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=-\frac{1}{2} a_{n}+\frac{1}{3^{n}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Find all values of $a_{1}$ such that $\left\{a_{n}\right\}$ is a monotonic sequence, i.e., $\left\{a_{n}\right\}$ is either an increasing sequence or a decreasing sequence.