The value of $\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$ [closed]

\begin{align*}\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\sum_{n=m+1}^\infty\frac{1}{2^n}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\cdot\frac{1}{2^m}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\cdot\frac{1}{6^l}\cdot\frac15\\&=\left(\frac{1}{1-\frac1{30}}-1\right)\cdot\frac15\\&=\frac{1}{29\cdot5}\end{align*} or rather $$\frac{1}{(5\cdot3\cdot2-1)(3\cdot2-1)(2-1)}$$

We have $$ \sum_{1\leq l<m<n}\frac{1}{5^l3^m2^n}=\sum_{l\geq1}\left(\frac{1}{5^l}\sum_{m>l}\left(\frac{1}{3^m}\sum_{n>m}\frac{1}{2^n}\right)\right). $$ Now $$ \sum_{n=m+1}^\infty\frac{1}{2^n}=2^{-m}, $$ so we can reduce the sum to $$ \sum_{l\geq1}\left(\frac{1}{5^l}\sum_{m>l}\left(\frac{1}{6}\right)^m\right). $$ Can you finish?

Outline:

You have $$ \sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n} = \sum_{n=1}^\infty\sum_{m=1}^{n-1}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell3^m2^n} = \sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} \tag{1} $$ Since $$ \sum_{\ell=1}^{m-1} \frac{1}{5^\ell} = \frac{1}{5}\cdot\frac{1-1/5^{m-1}}{1-1/5} = \frac{1-1/5^{m-1}}{4} = \frac{1-5/5^{m}}{4} $$ you can rewrite it as $$ \sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n} = \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1-5/5^{m}}{3^m} = \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m} - \frac{5}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{15^m}\,. $$ Since $\sum_{m=1}^{n-1}\frac{1}{3^m}$ and $\sum_{m=1}^{n-1}\frac{1}{15^m}$ can be computed as [...], you can rewrite [...]

Can you continue?