For all values of A, B, C and P, Q, R $\left|\begin{array}{ccc}\cos (A - P)& \cos (A-Q)& \cos (A-R)\\ \cos (B-P)& \cos (B-Q)& \cos (B-R)\\ \cos (C-P)& \cos (C-Q)& \cos (C-R)\end{array}\right| =$

$\left|\begin{array}{ccc}\cos (\mathrm{A} - \mathrm{P})& \cos (\mathrm{A}-\mathrm{Q})& \cos (\mathrm{A}-\mathrm{R})\\ \cos (\mathrm{B}-\mathrm{P})& \cos (\mathrm{B}-\mathrm{Q})& \cos (\mathrm{B}-\mathrm{R})\\ \cos (\mathrm{C}-\mathrm{P})& \cos (\mathrm{C}-\mathrm{Q})& \cos (\mathrm{C}-\mathrm{R})\end{array}\right|$

The given determinant is the product of 

$$\begin{gathered} =\left|\begin{array}{ccc}\cos (\mathrm{A} - \mathrm{P})& \cos (\mathrm{A}-\mathrm{Q})& \cos (\mathrm{A}-\mathrm{R})\\ \cos (\mathrm{B}-\mathrm{P})& \cos (\mathrm{B}-\mathrm{Q})& \cos (\mathrm{B}-\mathrm{R})\\ \cos (\mathrm{C}-\mathrm{P})& \cos (\mathrm{C}-\mathrm{Q})& \cos (\mathrm{C}-\mathrm{R})\end{array}\right| \\ \begin{array}{rcl} & =& \end{array}\begin{array}{rcl} & & \left|\begin{array}{ccc}\cos \mathrm{A}& \sin \mathrm{A}& 0\\ \cos \mathrm{B}& \sin \mathrm{B}& 0\\ \cos \mathrm{C}& \sin \mathrm{C}& 0\end{array}\right|\end{array}\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \left|\begin{array}{ccc}\cos \mathrm{P}& \cos \mathrm{Q}& \cos \mathrm{R}\\ \sin \mathrm{P}& \sin \mathrm{Q}& \sin \mathrm{R}\\ 0& 0& 0\end{array}\right|\end{array}\begin{array}{rcl} & & \end{array} \end{gathered}$$

$\begin{array}{rcl}{∆}_1& =& \left|\begin{array}{ccc}\cos \mathrm{A}& \sin \mathrm{A}& 0\\ \cos \mathrm{B}& \sin \mathrm{B}& 0\\ \cos \mathrm{C}& \sin \mathrm{C}& 0\end{array}\right| \mathrm{and} {∆}_2=\left|\begin{array}{ccc}\cos \mathrm{P}& \cos \mathrm{Q}& \cos \mathrm{R}\\ \sin \mathrm{P}& \sin \mathrm{Q}& \sin \mathrm{R}\\ 0& 0& 0\end{array}\right|\\ \therefore {∆}_1& =& 0, {∆}_2=0\end{array}$

and hence  ${∆}_1·{∆}_2=0$