A regular tetrahedron has vertices at the points
$$A\left(0, 0, \frac{2}{\sqrt{3}}\sqrt{6}\right), \quad B\left(\frac{2}{\sqrt{3}}\sqrt{3}, 0, 0\right), \quad C\left(-\frac{1}{3}\sqrt{3}, 1, 0\right), \quad D\left(-\frac{1}{3}\sqrt{3}, -1, 0\right).$$
- Obtain the equation of the face \(ABC\) in the form
$$x + \sqrt{3}y + \left(\frac{1}{2}\sqrt{2}\right)z = \frac{2}{3}\sqrt{3}.$$ [5]
(Answers which only verify the given equation will not receive full credit.)
- Give a geometrical reason why the equation of the face \(ABD\) can be expressed as
$$x - \sqrt{3}y + \left(\frac{1}{2}\sqrt{2}\right)z = \frac{2}{3}\sqrt{3}.$$ [2]
- Hence find the cosine of the angle between two faces of the tetrahedron. [4]