Let $\Delta=\begin{vmatrix} (a^x+a^{-x})^2 & (a^x-a^{-x})^2 & 1 \\ (a^y+a^{-y})^2 & (a^y-a^{-y})^2 & 1 \\ (a^z+a^{-z})^2 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Apply $C_1 \to C_1-C_2$
$(a+b)^2-(a-b)^2=4ab)$
$\Delta=\begin{vmatrix} 4 & (a^x-a^{-x})^2 & 1 \\ 4 & (a^y-a^{-y})^2 & 1 \\ 4 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Take 4 as the common factor from $C_1$
$\Delta=4\begin{vmatrix} 1 & (a^x-a^{-x})^2 & 1 \\ 1 & (a^y-a^{-y})^2 & 1 \\ 1 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Since two coloumns are identical, then the value of the determinant is 0
Therefore $\Delta =0$
Solution : option A is correct