Is this correct?
This group has Sylow subgroups of order 7, 11 and 19. It is easy to see that $n_7=1$. Also $n_{11}=1$ or $n_{11}=7*19$ and $n_{19}=1$ or $n_{19}=7*11$. To show that group of order 7*11*19 has normal subgroup of order 11 or order 19, it is enough to show that $n_{11}=7*19$ and $n_{19}=7*11$ can not hold in the same time, i.e. either one of $n_{11}$ and $n_{19}$ is 1. Suppose that none of them is 1. This means there are 7*19=133 cyclic 11-Sylow subgroups of order 11 and 7*11=77 cyclic 19-Sylow subgroups of order 19. Elements in 11-Sylow groups are all, except $e$, of order 11. Elements in 19-Sylow groups are all, except $e$, of order 19. This means there are 133*10=1330 elements of order 11 and 77*18=1386 elements of order 19, but group has $1463<1330+1386$ elements, which gives us a contradiction. So, either $n_{11}=1$ or $n_{19}=1$. This completes the proof.