11. As shown in Figure 3(a), the eight vertices of a cube are assigned values $a, b, c, d, e, f, g, h$, respectively. Then, the arithmetic mean of the values of the three adjacent vertices of each vertex, denoted as $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}, \bar{g}, \bar{h}$, is placed at the corresponding vertex of another cube, as shown in Figure 3(b).

If $\bar{a}=9, \bar{b}=8, \bar{c}=11, \bar{d}=10, \bar{e}=13, \bar{f}=$ $12, \bar{g}=15, \bar{h}=14$, then the value of $a+g$ is
11.20.

According to the problem, we have
$$
\begin{array}{l}
\bar{a}=\frac{b+d+e}{3}, \bar{b}=\frac{a+c+f}{3}, \bar{c}=\frac{b+d+g}{3}, \\
\bar{d}=\frac{a+c+h}{3}, e=\frac{a+f+h}{3}, \bar{f}=\frac{b+e+g}{3}, \\
\bar{g}=\frac{c+f+h}{3}, \bar{h}=\frac{d+e+g}{3} .
\end{array}
$$

Then, $a=(\bar{b}+\bar{d}+\bar{e})-2 \bar{g}, g=(\bar{c}+f+\bar{h})-2 \bar{a}$.
Thus, $a+g=(\bar{b}+\bar{c}+\bar{d}+\ddot{e}+f+\bar{h})-2(\vec{a}+\bar{g})$.
Given $\bar{a}=9, \bar{b}=8, \bar{c}=11, \bar{d}=10, e=13, \bar{f}=12, \bar{g}=$ $15, \bar{h}=14$, therefore, $a+g=20$.