For a triangle with side lengths $ a, $ $ b, $ $ c, $ let $ s = \frac{a + b + c}{2}, $ and let \begin{align*} x &= s - a = \frac{-a + b + c}{2}, \\ y &= s - b = \frac{a - b + c}{2}, \\ z &= s - c = \frac{a + b - c}{2}.\end{align*}By the Triangle Inequality, $ x, $ $ y, $ and $ z $ are all positive. (This technique is often referred to as the Ravi Substitution.) Note that \begin{align*} a &= y + z, \\ b &= x + z, \\ c &= x + y.\end{align*}If $ s $ is even, then $ x, $ $ y, $ and $ z $ are all positive integers. So, we can set $ x = i, $ $ y = j, $ and $ z = k, $ which gives us the parameterization $ (a,b,c) = (j + k, i + k, i + j) $. If $ s $ is odd, then $ x, $ $ y, $ and $ z $ are all of the form $ n - \frac{1}{2}, $ where $ n $ is a positive integer. So, we can set $ x = i - \frac{1}{2}, $ $ y = j - \frac{1}{2}, $ and $ z = k - \frac{1}{2} $. This gives us the parameterization $ (a,b,c) = (j + k - 1, i + k - 1, i + j - 1) $. Thus, our sum is \begin{align*} \sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c} &= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{2^{j + k - 1}}{3^{i + k - 1} 5^{i + j - 1}} \right) \\ &= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{15}{2} \cdot \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \right) \\ &= \frac{17}{2} \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \\ &= \frac{17}{2} \sum_{i = 1}^\infty \frac{1}{15^i} \sum_{j = 1}^\infty \left( \frac{2}{5} \right)^j \sum_{k = 1}^\infty \left( \frac{2}{3} \right)^k \\ &= \frac{17}{2} \cdot \frac{1/15}{1 - 1/15} \cdot \frac{2/5}{1 - 2/5} \cdot \frac{2/3}{1 - 2/3} \\ &= \boxed{\frac{17}{21}}.\end{align*}