I have the following character table. Note I assume that $\chi_i$'s are all irreducible.
$$ \begin{array}{|c|c|c|c|c|} \hline & C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline \chi_0 & 1 &1 & 1& 1&1 \\ \hline \chi_1 & 1 & \zeta_3^2 & \zeta_3 & 1 & 1 \\ \hline \chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline \chi_3 & 3 & 0 & 0 & \zeta_7^3+\zeta_7^4+\zeta_7^6 & \zeta_7+\zeta_7^0+\zeta_7^5 \\ \hline \chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline \end{array} $$
I would like the show that there is a normal subgroup of order $7$.
Now I would I am looking at the last two columns and thought that for there to be normal subgroup of order $7$ they would all need to equal $1$ because surely,
$$\text{Normal subgroup}=\bigcap_{\chi_i} \{ g \in G: |\chi_i(g)|=1 \} $$
Is this a correct expression, if not, what is the correct way of finding normal subgroups?