jmc

For a triangle with side lengths $a,$ $b,$ $c,$ let $s=\frac{a+b+c}{2},$ and let $$\begin{array}{rl}x& =s-a=\frac{-a+b+c}{2},\\ y& =s-b=\frac{a-b+c}{2},\\ z& =s-c=\frac{a+b-c}{2}.\end{array}$$By the Triangle Inequality, $x,$ $y,$ and $z$ are all positive. (This technique is often referred to as the Ravi Substitution.) Note that $$\begin{array}{rl}a& =y+z,\\ b& =x+z,\\ c& =x+y.\end{array}$$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 $$\begin{array}{rl}\sum _{(a,b,c)\in T}\frac{2^a}{3^b{5}^c}& =\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\left(\frac{2^{j+k}}{3^{i+k}{5}^{i+j}}+\frac{2^{j+k-1}}{3^{i+k-1}{5}^{i+j-1}}\right)\\ & =\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\left(\frac{2^{j+k}}{3^{i+k}{5}^{i+j}}+\frac{15}{2}\cdot \frac{2^{j+k}}{3^{i+k}{5}^{i+j}}\right)\\ & =\frac{17}{2}\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\frac{2^{j+k}}{3^{i+k}{5}^{i+j}}\\ & =\frac{17}{2}\sum _{i=1}^{\infty }\frac{1}{15^i}\sum _{j=1}^{\infty }{\left(\frac{2}{5}\right)}^j\sum _{k=1}^{\infty }{\left(\frac{2}{3}\right)}^k\\ & =\frac{17}{2}\cdot \frac{1/15}{1-1/15}\cdot \frac{2/5}{1-2/5}\cdot \frac{2/3}{1-2/3}\\ & =\boxed{\frac{17}{21}}.\end{array}$$