Let $p$ be the smallest prime divisor of $n$. I'll prove $p=17$. Notice that $6^{2n}\equiv 11^{2n}\pmod{p}$ and $6^{p-1}\equiv 11^{p-1}\pmod{p}$. So, $6^t\equiv 11^t\pmod{p}$, where $t=(2n,p-1)=2$ (clearly $n$ is odd). So, $p\mid 11^2-6^2=85=5\cdot 17$. Since $6^n+11^n\equiv 2\pmod{5}$, we obtain $p=17$. Now let $n=17^i n'$ where $17\nmid n'$. It is evident that $n'\mid n^{100}+6^n+11^n$, so it suffices to show $v_17(n^{100}+6^n+11^n)=i+1$. Note that by LTE, we have $v_{17}(6^n+11^n) = 1+v_17(n)=1+i$. Since $v_{17}(n^{100})=100i\gg i+1$, we obtain that $v_17(n^{100}+6^n+11^n)=i+1$, as claimed.