The position vectors of points A, B and C are such that OA = 3i − 5j − 4k, OB = j + 8k and OC = -2i + 5j + 16k. Show that the points A, B and C are collinear.

$$\vec{OA} = \begin{pmatrix} 3 \\ -5 \\ -4 \end{pmatrix}$$ $$\vec{OB} = \begin{pmatrix} 0 \\ 1 \\ 8 \end{pmatrix}$$ $$\vec{OC} = \begin{pmatrix} -2 \\ 5 \\ 16 \end{pmatrix}$$

We need to find a scalar $k$ such that $\vec{AB} = k \vec{BC}$.

$$\begin{pmatrix} -3 \\ 6 \\ 12 \end{pmatrix} = k \begin{pmatrix} -2 \\ 4 \\ 8 \end{pmatrix}$$

Comparing the components:
For the x-component: $-3 = k(-2) \implies k = \frac{-3}{-2} = \frac{3}{2}$
For the y-component: $6 = k(4) \implies k = \frac{6}{4} = \frac{3}{2}$
For the z-component: $12 = k(8) \implies k = \frac{12}{8} = \frac{3}{2}$
Since $k = \frac{3}{2}$ is consistent for all components, $\vec{AB} = \frac{3}{2} \vec{BC}$.
This shows that vectors $\vec{AB}$ and $\vec{BC}$ are parallel. Since they share a common point B, the points A, B, and C are collinear.

$$\boxed{\text{Points A, B, and C are collinear.}}$$

15. Given that $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$, simplify $\frac{1}{\sin 75^\circ}$.

Use the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ in the denominator.

$$= \frac{4(\sqrt{6} - \sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2}$$ $$= \frac{4(\sqrt{6} - \sqrt{2})}{6 - 2}$$ $$= \frac{4(\sqrt{6} - \sqrt{2})}{4}$$

$$= \sqrt{6} - \sqrt{2}$$

The simplified expression is:

$$\boxed{\sqrt{6} - \sqrt{2}}$$