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# For all values of $A,B,C$ and $P,Q,R$ the value of the determinant $(x+a)^3\small\begin{vmatrix}\cos(A-P)&\cos (A-Q)&\cos(A-R)\\\cos(B-P)&\cos(B-Q)&\cos(B-R)\\\cos(C-P)&\cos(C-Q)&\cos(C-R)\end{vmatrix}$ is

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;None\;of\;these$

We have
$\begin{vmatrix}\cos(A-P)&\cos (A-Q)&\cos(A-R)\\\cos(B-P)&\cos(B-Q)&\cos(B-R)\\\cos(C-P)&\cos(C-Q)&\cos(C-R)\end{vmatrix}=\begin{vmatrix}\cos A&\sin A&0\\\cos B&\sin B&0\\\cos C&\sin C&0\end{vmatrix}\times \begin{vmatrix} \cos P&\sin P&0\\\cos Q&\sin Q&0\\\cos R&\sin R&0\end{vmatrix}$
$\Rightarrow 0\times 0$
$\Rightarrow 0$
Hence (b) is the correct answer.