= Solution
Choose the <unit vector> $\mathbf p=(1,-1,0)/\sqrt2$, which is perpendicular to $\mathbf v$. Then $M\mathbf p=(0,1,-1)/\sqrt2$. Their dot product is $-1/2$, and the <rotation matrix> preserves their norms, so
$$
\boxed{\angle(\mathbf p,M\mathbf p)=\arccos(-1/2)=\frac{2\pi}{3}.}
$$
The same unsigned angle holds for every nonzero vector perpendicular to the axis: this is the rotation plane of a three-dimensional proper <rotation>. Alternatively, $\operatorname{tr}M=0=1+2\cos\chi$ gives the same rotation angle $\chi=2\pi/3$.
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