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Question

Let $\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$ and $\mathrm{R}(7,3,2)$ be the vertices of $\triangle \mathrm{PQR}$. Then, the angle $\angle \mathrm{QPR}$ is

$\cos ^{-1}\left(\frac{7}{18}\right)$
$\frac{\pi}{6}$
$\cos ^{-1}\left(\frac{1}{18}\right)$
$\frac{\pi}{3}$

Solution

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