7. Let the set $T=\left\{(a, b, c) \mid a, b, c \in \mathbf{N}^{*}\right.$, and $a, b, c$ can form the side lengths of some triangle $\}$. Then the value of $\sum_{(a, b, c) \in T} \frac{2^{a}}{3^{b} \cdot 5^{c}}$ is $\qquad$.
It is known that $a, b, c$ are the lengths of the three sides of a triangle if and only if there exist positive numbers $x, y, z$ satisfying
$$
a=\frac{x+y}{2}, b=\frac{y+z}{2}, c=\frac{z+x}{2} \Rightarrow x=a+c-b, y=a+b-c, z=b+c-a \text {, }
$$

Thus, $x, y, z$ are positive integers with the same parity. Let $x=2 p-1, y=2 q-1, z=2 r-1$ or $x=2 p, y=2 q, z=2 r$, where $p, q, r \in \mathbf{N}^{*}$.
$$
\begin{array}{l}
\text { Then } \sum_{(a, b, c) \in T} \frac{2^{a}}{3^{b} \cdot 5^{c}}=\sum_{p, q, r \in \mathbf{N}^{*}} \frac{2^{p+q-1}}{3^{q+r-1} \cdot 5^{p+r-1}}+\sum_{p, q, r \in \mathbf{N}^{*}} \frac{2^{p+q}}{3^{q+r} \cdot 5^{p+r}} \\
=\left(\frac{2^{-1}}{3^{-1} \cdot 5^{-1}}+1\right) \sum_{p, q, r \in \mathbf{N}^{*}} \frac{2^{p+q}}{3^{q+r} \cdot 5^{p+r}}=\frac{17}{2} \sum_{p, q, r \in \mathbf{N}^{*}}\left[\left(\frac{2}{5}\right)^{p} \cdot\left(\frac{2}{3}\right)^{q} \cdot\left(\frac{1}{15}\right)^{r}\right] \\
=\frac{17}{2} \cdot \sum_{p \in \mathbf{N}^{*}}\left(\frac{2}{5}\right)^{p} \cdot \sum_{q \in \mathbf{N}^{*}}\left(\frac{2}{3}\right)^{q} \cdot \sum_{r \in \mathbf{N}^{*}}\left(\frac{1}{15}\right)^{r}=\frac{17}{2} \cdot \frac{\frac{2}{5}}{1-\frac{2}{5}} \cdot \frac{\frac{2}{3}}{1-\frac{2}{3}} \cdot \frac{\frac{1}{15}}{1-\frac{1}{15}} \\
=\frac{17}{2} \cdot \frac{2}{3} \cdot 2 \cdot \frac{1}{14}=\frac{17}{21} .
\end{array}
$$