1. We need to evaluate $I=\int_{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2x^{14}+3x^{7}+6\right)^{1/7}dx.$

2. First, factor the polynomial: $x^{21}+x^{14}+x^7=x^7(x^{14}+x^7+1).$

Let $t=2x^{14}+3x^7+6.$ Then $\frac{dt}{dx}=28x^{13}+21x^6=7x^6(4x^7+3).$ This does not directly match the given factor, so we try a better substitution.

3. Let $u=x^7.$ Then $du=7x^6dx,\qquad dx=\frac{1}{7}u^{-6/7}du.$ Also, $x^{21}=u^3,\quad x^{14}=u^2,\quad x^7=u.$ Hence $x^{21}+x^{14}+x^7=u^3+u^2+u=u(u^2+u+1).$ And $2x^{14}+3x^7+6=2u^2+3u+6.$ So $I=\frac17\int_0^1 (u^3+u^2+u)(2u^2+3u+6)^{1/7}u^{-6/7}du.$ This becomes messy, so instead we look for a derivative pattern in terms of $x^7$.

4. Observe that if $F(x)=\left(2x^{14}+3x^7+6\right)^{8/7},$ then $F'(x)=\frac87\left(2x^{14}+3x^7+6\right)^{1/7}(28x^{13}+21x^6).$ So $F'(x)=8\left(2x^{14}+3x^7+6\right)^{1/7}(4x^{13}+3x^6).$ Still not matching.

5. Now use the substitution $u=x^7+\frac{1}{x^7}?$ No, that is not suitable on $[0,1]$.

Instead notice: $x^{21}+x^{14}+x^7=x^7(x^{14}+x^7+1).$ Let $t=2x^{14}+3x^7+6.$ Then $dt=7x^6(4x^7+3)dx.$ This suggests writing the integrand in terms of $x^7$ and then using $y=x^7$.

6. Put $y=x^7\Rightarrow x=y^{1/7},\quad dx=\frac17 y^{-6/7}dy.$ Then

That is $I=\frac17\int_0^1 y^{1/7}(y^2+y+1)(2y^2+3y+6)^{1/7}dy.$ Now note that $2y^2+3y+6=(y+2)(2y+3),$ not especially useful.

7. Try another substitution: $t=(2x^{14}+3x^7+6)^{8/7}.$ But a cleaner route is to observe $x^{21}+x^{14}+x^7=x^7\cdot \frac{(4x^7+3)x^7+?}{?}$ This is cumbersome.

Let us instead set $u=2x^{14}+3x^7+6.$ Write $y=x^7$, so $u=2y^2+3y+6$. Then $du=(4y+3)dy.$ Also, $y^3+y^2+y=y(y^2+y+1).$ Now divide: $y^3+y^2+y=\frac14(4y+3)(y^2+\cdots)$ But this again is not direct.

8. So try to express the integral as an exact derivative: Assume $\frac{d}{dx}\left[x^a(2x^{14}+3x^7+6)^{8/7}\right]$ may generate the integrand.

Take $G(x)=x^8(2x^{14}+3x^7+6)^{8/7}.$ Then \begin{align*} G'(x)&=8x^7(2x^{14}+3x^7+6)^{8/7} +x^8\cdot \frac87(2x^{14}+3x^7+6)^{1/7}(28x^{13}+21x^6)\ &=8x^7(2x^{14}+3x^7+6)^{1/7}\Big[(2x^{14}+3x^7+6)+x^7(4x^7+3)\Big]\ &=8x^7(2x^{14}+3x^7+6)^{1/7}(6x^{14}+6x^7+6)\ &=48(x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}. \end{align*} This matches the integrand perfectly.

Hence $I=\frac1{48}\left[x^8(2x^{14}+3x^7+6)^{8/7}\right]_0^1.$

9. Evaluate at the limits: At $x=1$, $x^8(2x^{14}+3x^7+6)^{8/7}=1\cdot(2+3+6)^{8/7}=11^{8/7}.$ At $x=0$, $x^8(2x^{14}+3x^7+6)^{8/7}=0.$ Therefore $I=\frac{1}{48}11^{8/7}.$

10. Compare with $I=\frac1l(11)^{m/n}.$ So, $l=48,\quad m=8,\quad n=7.$ Thus $l+m+n=48+8+7=63.$

Therefore the required integer is $\boxed{63}.$