= Solution
{c}
Use the corrected decomposition
$$
y_{ij}-w_j=(y_{ij}-\mu_j)+(\mu_j-w_j).
$$
The hint's minus sign before $(\mu_j-w_j)$ is a typographical error. Squaring, summing, and using $\sum_i(y_{ij}-\mu_j)=0$ for each coordinate gives the <centroid sum-of-squares identity>
$$
\sum_{i,j}(y_{ij}-w_j)^2
=\sum_{i,j}(y_{ij}-\mu_j)^2
+2\sum_j(\mu_j-w_j)\underbrace{\sum_i(y_{ij}-\mu_j)}_{0}
+m\sum_j(\mu_j-w_j)^2.
$$
Since $m>0$ and $w\ne\mu$, the final term is strictly positive. Hence
$$
\boxed{\sum_{i,j}(y_{ij}-\mu_j)^2<\sum_{i,j}(y_{ij}-w_j)^2.}
$$
Thus the <centroid> uniquely minimizes the sum of squared <Euclidean distances> to the observations.
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