(i) First, find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3i + 2j + 8k) - (i - 2j + 4k) = 2i + 4j + 4k\)
\(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = (3i + 2j + 8k) - (-i - 2j + 10k) = 4i + 4j - 2k\)
Calculate the scalar product \(\overrightarrow{AB} \cdot \overrightarrow{CB}\):
\(\overrightarrow{AB} \cdot \overrightarrow{CB} = (2i + 4j + 4k) \cdot (4i + 4j - 2k) = 2 \times 4 + 4 \times 4 + 4 \times (-2) = 8 + 16 - 8 = 16\)
Find the magnitudes of \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):
\(|\overrightarrow{AB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\)
\(|\overrightarrow{CB}| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\)
Use the scalar product to find \(\cos \theta\):
\(\overrightarrow{AB} \cdot \overrightarrow{CB} = |\overrightarrow{AB}| \cdot |\overrightarrow{CB}| \cdot \cos \theta\)
\(16 = 6 \times 6 \times \cos \theta\)
\(\cos \theta = \frac{16}{36} = \frac{4}{9}\)
\(\theta = \cos^{-1}\left(\frac{4}{9}\right) \approx 63.6^\circ\)
(ii) Find the lengths of \(AB\), \(BC\), and \(CA\):
\(AB = |\overrightarrow{AB}| = 6\)
\(BC = |\overrightarrow{CB}| = 6\)
\(CA = |\overrightarrow{CA}| = \sqrt{(-2)^2 + 0^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}\)
Perimeter of \(\triangle ABC = AB + BC + CA = 6 + 6 + \sqrt{40} \approx 18.32\)
