= Solution
{c}
The <straight line> $2x=y=z$ has <direction vector> $(1,2,2)$, so take the <unit vector> $\mathbf n=(1,2,2)^T/3$. A <half-turn> has $\cos\pi=-1$ and $\sin\pi=0$. The preceding <rotation> formula becomes
$$
R(\mathbf v)=2(\mathbf v\cdot\mathbf n)\mathbf n-\mathbf v.
$$
Thus the <rotation matrix> in the <standard basis> is
$$
\boxed{2\mathbf n\mathbf n^T-I=\frac19\begin{pmatrix}-7&4&4\\4&-1&8\\4&8&-1\end{pmatrix}.}
$$
It fixes $(1,2,2)^T$ and negates every <vector> in its <orthogonal complement>, which also checks the <half-turn> geometrically.
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