Surface area of polyhedron

7 (In this question, the notation \(\triangle A B C\) denotes the area of the triangle \(A B C\).)
The points \(P , Q\) and \(R\) have position vectors \(p \mathbf { i } , q \mathbf { j }\) and \(r \mathbf { k }\) respectively, relative to the origin \(O\), where \(p , q\) and \(r\) are positive. The points \(O , P , Q\) and \(R\) are joined to form a tetrahedron.

1. Draw a sketch of the tetrahedron and write down the values of \(\triangle O P Q , \triangle O Q R\) and \(\triangle O R P\).
2. Use the definition of the vector product to show that \(\frac { 1 } { 2 } | \overrightarrow { R P } \times \overrightarrow { R Q } | = \Delta P Q R\).
3. Show that \(( \triangle O P Q ) ^ { 2 } + ( \triangle O Q R ) ^ { 2 } + ( \triangle O R P ) ^ { 2 } = ( \triangle P Q R ) ^ { 2 }\).
4. Use de Moivre's theorem to express \(\cos 4 \theta\) as a polynomial in \(\cos \theta\).
5. Hence prove that \(\cos 4 \theta \cos 2 \theta \equiv 16 \cos ^ { 6 } \theta - 24 \cos ^ { 4 } \theta + 10 \cos ^ { 2 } \theta - 1\).
6. Use part (ii) to show that the only roots of the equation \(\cos 4 \theta \cos 2 \theta = 1\) are \(\theta = n \pi\), where \(n\) is an integer.
7. Show that \(\cos 4 \theta \cos 2 \theta = - 1\) only when \(\cos \theta = 0\).

5 The tetrahedron \(T\), shown below, has vertices at \(O ( 0,0,0 ) , A ( 1,2,2 ) , B ( 2,1,2 )\) and \(C ( 2,2,1 )\). \includegraphics[max width=\textwidth, alt={}, center]{59fa1650-a296-471e-93b9-0988177cd89d-3_360_464_319_555} Diagram not drawn to scale Show that the surface area of \(T\) is \(\frac { 1 } { 2 } \sqrt { 3 } ( 1 + \sqrt { 51 } )\).