= Solution
In <suffix notation>, $x'_i=ax_i+b\epsilon_{ikj}n_kx_j$, so the <matrix> entries are
$$
\boxed{M_{ij}=a\delta_{ij}+b\epsilon_{ikj}n_k
=a\delta_{ij}-b\epsilon_{ijk}n_k.}
$$
Here $\delta$ is the <Kronecker delta> and $\epsilon$ the <Levi-Civita symbol>. The diagonal determines $a=2$. Comparing the off-diagonal entries with
$$
b[\mathbf n\times]=\begin{pmatrix}
0&-bn_3&bn_2\\bn_3&0&-bn_1\\-bn_2&bn_1&0
\end{pmatrix}
$$
gives $b\mathbf n=(1,2,2)$. As $\mathbf n$ is a <unit vector> and $b>0$, \b[$b=3$ and $\mathbf n=(1,2,2)/3$]. Thus
$$
\boxed{a=2,\qquad b=3,\qquad\mathbf n=\tfrac13(1,2,2).}
$$
In this example the transverse scaling is $\sqrt{13}$ and the axial scaling is two; the <rotation> angle is $\arctan(3/2)$.
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