If we are given a determinant \[\Delta =\left| \begin{align}
  & \begin{matrix}
   \cos \left( {{\alpha }_{1}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{1}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{1}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
 & \begin{matrix}
   \cos \left( {{\alpha }_{2}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{2}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{2}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
 & \begin{matrix}
   \cos \left( {{\alpha }_{3}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{3}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{3}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
\end{align} \right|\] then the value of \[\Delta \] equal to
(A) \[\cos {{\alpha }_{1}}\cos {{\alpha }_{2}}\cos {{\alpha }_{3}}\cos {{\beta }_{1}}\cos {{\beta }_{2}}\cos {{\beta }_{3}}\]
(B) \[\cos {{\alpha }_{1}}+\cos {{\alpha }_{2}}+\cos {{\alpha }_{3}}+\cos {{\beta }_{1}}+\cos {{\beta }_{2}}+\cos {{\beta }_{3}}\]
(C) \[\cos \left( {{\alpha }_{1}}-{{\beta }_{1}} \right)\cos \left( {{\alpha }_{2}}-{{\beta }_{2}} \right)\cos \left( {{\alpha }_{3}}-{{\beta }_{3}} \right)\]
(D) 0

Hint: Use the formula, \[\cos \left( A-B \right)=\cos A\cos B+ \sin A\sin B\] and expand the terms \[\cos \left( {{\alpha }_{1}}-{{\beta }_{1}} \right)\] , \[\cos \left( {{\alpha }_{1}}-{{\beta }_{2}} \right)\] , \[\cos \left( {{\alpha }_{1}}-{{\beta }_{3}} \right)\] , \[\cos \left( {{\alpha }_{2}}-{{\beta }_{1}} \right)\] , \[\cos \left( {{\alpha }_{2}}-{{\beta }_{2}} \right)\] , \[\cos \left( {{\alpha }_{2}}-{{\beta }_{3}} \right)\] , \[\cos \left( {{\alpha }_{3}}-{{\beta }_{1}} \right)\] , \[\cos \left( {{\alpha }_{3}}-{{\beta }_{2}} \right)\] , and \[\cos \left( {{\alpha }_{3}}-{{\beta }_{3}} \right)\] . Now, split it as the multiplication of two determinants. Finally, use the property that the determinant value of a matrix is equal to zero if all the elements of a row or a column is equal to zero and get the answer.

Complete step-by-step solution
According to the question, we are given an expression for \[\Delta \] and we are asked to find the value of
\[\Delta \] .
\[\Delta =\left| \begin{align}
  & \begin{matrix}
   \cos \left( {{\alpha }_{1}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{1}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{1}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
 & \begin{matrix}
   \cos \left( {{\alpha }_{2}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{2}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{2}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
 & \begin{matrix}
   \cos \left( {{\alpha }_{3}}-{{\beta }_{1}} \right) & \cos \left( {{\alpha }_{3}}-{{\beta }_{2}} \right) & \cos \left( {{\alpha }_{3}}-{{\beta }_{3}} \right) \\
\end{matrix} \\
\end{align} \right|\] ……………………………………………(1)
We know the formula that \[\cos \left( A-B \right)=\cos A\cos B+ \sin A\sin B\] ……………………………………….(2)
Replacing A and B by \[{{\alpha }_{1}}\] and \[{{\beta }_{1}}\] in equation (2), we get
 \[\cos \left( {{\alpha }_{1}}-{{\beta }_{1}} \right)=\cos {{\alpha }_{1}}\cos {{\beta }_{1}}+\sin {{\alpha }_{1}}\sin {{\beta }_{1}}\] ……………………………………….(3)
Similarly,
\[\cos \left( {{\alpha }_{1}}-{{\beta }_{2}} \right)=\cos {{\alpha }_{1}}\cos {{\beta }_{2}}+\sin {{\alpha }_{1}}\sin {{\beta }_{2}}\] ………………………………………….(4)
\[\cos \left( {{\alpha }_{1}}-{{\beta }_{3}} \right)=\cos {{\alpha }_{1}}\cos {{\beta }_{3}}+\sin {{\alpha }_{1}}\sin {{\beta }_{3}}\] ……………………………………………(5)
\[\cos \left( {{\alpha }_{2}}-{{\beta }_{1}} \right)=\cos {{\alpha }_{2}}\cos {{\beta }_{1}}+\sin {{\alpha }_{2}}\sin {{\beta }_{1}}\] ……………………………………………(6)
\[\cos \left( {{\alpha }_{2}}-{{\beta }_{2}} \right)=\cos {{\alpha }_{2}}\cos {{\beta }_{2}}+\sin {{\alpha }_{2}}\sin {{\beta }_{2}}\] …………………………………………..(7)
\[\cos \left( {{\alpha }_{2}}-{{\beta }_{3}} \right)=\cos {{\alpha }_{2}}\cos {{\beta }_{3}}+\sin {{\alpha }_{2}}\sin {{\beta }_{3}}\] …………………………………………….(8)
\[\cos \left( {{\alpha }_{3}}-{{\beta }_{1}} \right)=\cos {{\alpha }_{3}}\cos {{\beta }_{1}}+\sin {{\alpha }_{3}}\sin {{\beta }_{1}}\] ………………………………………………..(9)
\[\cos \left( {{\alpha }_{3}}-{{\beta }_{2}} \right)=\cos {{\alpha }_{3}}\cos {{\beta }_{2}}+\sin {{\alpha }_{3}}\sin {{\beta }_{2}}\] ………………………………………………(10)
\[\cos \left( {{\alpha }_{3}}-{{\beta }_{3}} \right)=\cos {{\alpha }_{3}}\cos {{\beta }_{3}}+\sin {{\alpha }_{3}}\sin {{\beta }_{3}}\] …………………………………….(11)
Now, using the above equations and on simplifying equation (1), we get
\[\Delta =\left| \begin{align}
  & \begin{matrix}
   \cos {{\alpha }_{1}}\cos {{\beta }_{1}}+\sin {{\alpha }_{1}}\sin {{\beta }_{1}} & \cos {{\alpha }_{1}}\cos {{\beta }_{2}}+\sin {{\alpha }_{1}}\sin {{\beta }_{2}} & \cos {{\alpha }_{1}}\cos {{\beta }_{3}}+\sin {{\alpha }_{1}}\sin {{\beta }_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   \cos {{\alpha }_{2}}\cos {{\beta }_{1}}+\sin {{\alpha }_{2}}\sin {{\beta }_{1}} & \cos {{\alpha }_{2}}\cos {{\beta }_{2}}+\sin {{\alpha }_{2}}\sin {{\beta }_{2}} & \cos {{\alpha }_{2}}\cos {{\beta }_{3}}+\sin {{\alpha }_{2}}\sin {{\beta }_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   \cos {{\alpha }_{3}}\cos {{\beta }_{1}}+\sin {{\alpha }_{3}}\sin {{\beta }_{1}} & \cos {{\alpha }_{3}}\cos {{\beta }_{2}}+\sin {{\alpha }_{3}}\sin {{\beta }_{2}} & \cos {{\alpha }_{3}}\cos {{\beta }_{3}}+\sin {{\alpha }_{3}}\sin {{\beta }_{3}} \\
\end{matrix} \\
\end{align} \right|\] ……………………………………………….(12)
We can see that the above matrix is somewhat complex which needs to be simplified more.
Let us break the above into the multiplication of two determinant matrices.
On breaking the matrix shown in equation (12), we get
\[\Delta =\left| \begin{align}
  & \begin{matrix}
   \cos {{\alpha }_{1}} & \sin {{\alpha }_{1}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   \cos {{\alpha }_{2}} & \sin {{\alpha }_{2}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   \cos {{\alpha }_{3}} & \sin {{\alpha }_{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\]$\times$ \[\left| \begin{align}
  & \begin{matrix}
   \cos {{\beta }_{1}} & \cos {{\beta }_{2}} & \cos {{\beta }_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   \sin {{\beta }_{1}} & \,\sin {{\beta }_{2}} & \,\,\sin {{\beta }_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   \,\,\,\,\,\,0 & \,\,\,\,\,\,\,\,0 & \,\,\,\,\,\,\,\,\,\,0 \\
\end{matrix} \\
\end{align} \right|\] …………………………………………(13)
We know the property that the determinant value of matrix is equal to zero if all the elements of a row or a column is equal to zero ……………………………………(14)
Now, using the property shown in equation (14) and on simplifying equation (13), we get
\[\Delta =0\times 0=0\] ……………………………………….(15)
Therefore, the value of \[\Delta \] is equal to zero.
Hence, the correct option is (D).

Note: Whenever this type of question appears where we are given a complex determinant then, try to break the given matrix expression as the product of two determinants. Now, use the property that the determinant value of the matrix is equal to zero if all the elements of a row or a column are equal to zero.