Question:

<p><strong>Step 1:</strong> Given: $\cot^{-1}\left(\frac{n^2 - 10n + 21}{\pi}\right) > \frac{\pi}{6}$</p><p><strong>Step 2:</strong> Since $\cot x$ is decreasing on $[0, \pi]$, we have:</p><p>$$\cot\left(\cot^{-1}\left(\frac{n^2 - 10n + 21}{\pi}\right)\right) < \cot\frac{\pi}{6}$$</p><p>$$\frac{n^2 - 10n + 21}{\pi} < \sqrt{3}$$</p><p><strong>Step 3:</strong> This gives:</p><p>$$n^2 - 10n + 21 < \pi\sqrt{3}$$</p><p>$$n^2 - 10n + 21 < 5.44$$ (approximately)</p><p>$$n^2 - 10n + 15.56 < 0$$</p><p><strong>Step 4:</strong> Solving: $n < 5.5$ (approximately)</p><p>Therefore, the maximum value of $n$ is $\boxed{5}$</p>