= Solution
Start from the three-index transformation law and contract the first two Cartesian indices:
$$
\begin{aligned}
C'_k=\sum_iS'_{iik}&=\sum_{i,a,b,c}Q_{ia}Q_{ib}Q_{kc}S_{abc}\\
&=\sum_{a,b,c}\delta_{ab}Q_{kc}S_{abc}=\sum_cQ_{kc}\sum_aS_{aac}=Q_{kc}C_c.
\end{aligned}
$$
Orthogonality supplies $\sum_iQ_{ia}Q_{ib}=\delta_{ab}$, where $\delta$ is the <Kronecker delta>. The remaining factor is exactly the <vector> transformation rule. Therefore \b[$C_k$ is a vector].
This <tensor contraction> removes two indices by pairing them through the Euclidean metric. In general nonorthonormal coordinates, contraction must use the appropriate metric or pair a covariant and a contravariant index; the bare repeated-lower-index sum relies on Cartesian orthonormal coordinates.
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