If $$\int_0^1 x^{21} + x^{14} + x^7 2x^{14} + 3x^7 + 6^{1/7} dx = \frac{1}{l}(11)^{m/n}$$ where $$l, m, n \in \mathbb{N}$$, $$m$$ and $$n$$ are co-prime then $$l + m + n$$ is equal to _____.

We need to evaluate $$\int_0^1 (x^{21} + x^{14} + x^7)(2x^{14} + 3x^7 + 6)^{1/7}\,dx$$. To simplify this, set $$t = x^7$$ so that $$dt = 7x^6\,dx$$, giving $$dx = \frac{dt}{7x^6} = \frac{dt}{7t^{6/7}}$$.

Since $$x^{21} = t^3$$, $$x^{14} = t^2$$, and $$x^7 = t$$, substituting into the integral transforms it into $$I = \int_0^1 (t^3 + t^2 + t)(2t^2 + 3t + 6)^{1/7} \cdot \frac{dt}{7t^{6/7}} = \frac{1}{7}\int_0^1 t^{1/7}(t^2 + t + 1)(2t^2 + 3t + 6)^{1/7}\,dt\,. $$

Combining the seventh-root factors yields $$t^{1/7}(2t^2 + 3t + 6)^{1/7} = [t(2t^2 + 3t + 6)]^{1/7} = (2t^3 + 3t^2 + 6t)^{1/7}\,, $$ so that the integral becomes $$I = \frac{1}{7}\int_0^1 (t^2 + t + 1)(2t^3 + 3t^2 + 6t)^{1/7}\,dt\,. $$

Now let $$u = 2t^3 + 3t^2 + 6t\,. $$ Then $$\frac{du}{dt} = 6t^2 + 6t + 6 = 6(t^2 + t + 1)\,, $$ which implies $$(t^2 + t + 1)\,dt = \frac{du}{6}\,. $$ Substituting gives $$I = \frac{1}{7}\int_{t=0}^1 u^{1/7}\,\frac{du}{6} = \frac{1}{42}\int_{u=0}^{11} u^{1/7}\,du\,, $$ since when $$t = 0$$, $$u = 0$$ and when $$t = 1$$, $$u = 2 + 3 + 6 = 11$$.

Evaluating this yields $$I = \frac{1}{42}\left[\frac{u^{8/7}}{8/7}\right]_0^{11} = \frac{1}{42}\cdot\frac{7}{8}\cdot 11^{8/7} = \frac{1}{48}\cdot 11^{8/7}\,. $$

Comparing with the given form $$\frac{1}{l}(11)^{m/n}$$ shows that $$l = 48$$, $$m = 8$$, and $$n = 7$$ with $$\gcd(8,7)=1$$. Therefore $$l + m + n = 48 + 8 + 7 = 63$$.

The answer is $$\boxed{63}$$.