= Solution
The degree-sum formula for <irreducible characters> gives
$$
\boxed{|G|=1^2+1^2+1^2+3^2+3^2=21.}
$$
The number of <linear characters> is the order of the <abelianization>, so \b[$|G/G'|=3$] and $|G'|=7$. The intersection of the kernels of all <linear characters> is $G'$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed <linear character> have order three, so their orders are divisible by three. <Lagrange's theorem> leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three <irreducible characters>. Therefore
$$
\boxed{(|g_2|,|g_3|,|g_4|,|g_5|)=(3,3,7,7).}
$$
The center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an <abelian group>, again a contradiction. Thus a <centralizer> of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.
Column orthogonality says $\sum_i|\chi_i(g)|^2=|C_G(g)|$. For $g_2,g_3$, the three <linear characters> already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says $\sum_i\chi_i(1)\overline{\chi_i(g)}=0$ for $g\ne1$. At $g_4$ this gives $1+\alpha+\chi_5(g_4)=0$, so $\chi_5(g_4)=\beta$, using $\alpha+\beta=-1$. Similarly $\chi_5(g_5)=\alpha$. Thus the completed table is
$$
\boxed{\begin{array}{c|ccccc}
&e&g_2&g_3&g_4&g_5\\\hline
\text{class size}&1&7&7&3&3\\
\chi_1&1&1&1&1&1\\
\chi_2&1&\omega&\omega^2&1&1\\
\chi_3&1&\omega^2&\omega&1&1\\
\chi_4&3&0&0&\alpha&\beta\\
\chi_5&3&0&0&\beta&\alpha
\end{array}}
$$
Here $\beta=\overline\alpha$, $|\alpha|^2=|\beta|^2=2$ and $\alpha\beta=2$. These relations, together with $1+\omega+\omega^2=0$, verify all weighted row orthogonality relations and the <centralizer> column norms.
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