= Solution
{c}
First find the fixed <axis of rotation> by solving $(M-I)\mathbf n=0$. After removing a nonzero common factor, the <linear equations> are
$$
-2n_1+n_2-3n_3=0,\qquad -3n_1-2n_2-n_3=0,\qquad n_1+3n_2-2n_3=0.
$$
They give $n_2=n_3=-n_1$. Choose the following <unit vectors>:
$$
\mathbf p_1=\frac1{\sqrt3}(1,-1,-1)^T,\quad \mathbf p_2=\frac1{\sqrt2}(1,1,0)^T,\quad \mathbf p_3=\mathbf p_1\times\mathbf p_2=\frac1{\sqrt6}(1,-1,2)^T.
$$
Their <dot products> show that they form an <orthonormal basis>; hence $P=(\mathbf p_1\ \mathbf p_2\ \mathbf p_3)$ is an <orthogonal matrix>. Since $M$ is an <orthogonal matrix> fixing $\mathbf p_1$, it preserves the <orthogonal complement> of $\mathbf p_1$. Its restriction there has <determinant> $1$, so it is a <planar rotation>.
The <trace> gives $1+2\cos\theta=\operatorname{tr}M=9/7$, so $\cos\theta=1/7$. To determine the sign of the <sine>, rather than just its magnitude, calculate
$$
M\mathbf p_2=\frac1{7\sqrt2}(5,-3,8)^T,\qquad \mathbf p_3^TM\mathbf p_2=\frac{24}{7\sqrt{12}}=\frac{4\sqrt3}{7}>0.
$$
Thus a suitable <angle> and <change of basis> are
$$
\boxed{\theta=\arccos(1/7),\qquad P=\begin{pmatrix}1/\sqrt3&1/\sqrt2&1/\sqrt6\\-1/\sqrt3&1/\sqrt2&-1/\sqrt6\\-1/\sqrt3&0&2/\sqrt6\end{pmatrix}.}
$$
In fact direct <matrix multiplication> gives
$$
P^TMP=\begin{pmatrix}1&0&0\\0&1/7&-4\sqrt3/7\\0&4\sqrt3/7&1/7\end{pmatrix}=\operatorname{diag}(1,R_\theta).
$$
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