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Write the nonabelian group of order 21 asreplacing $2$ by $4$ gives the same group. Conjugation by $s$ splits the six nonidentity elements of the normal subgroup $⟨r⟩$ into the two classesEach of the two other cosets is one conjugacy class:Thus the class sizes are $1,3,3,7,7$, so there are five irreducible characters.

The abelianization is $C_3$, giving three linear characters. Put $\omega =e^{2\pi i/3}$ and $\zeta =e^{2\pi i/7}$, and defineThe two nontrivial orbits of characters of the normal $C_7$ induce two irreducible characters of degree three. Their values vanish outside $C_7$, and summing each orbit gives $\alpha ,\beta$ on the two nonidentity classes. The resulting character table isThe degree checkand character orthogonality relations verify that this is the complete table.