If $\cos 2\theta = 0$, then $\left| {\begin{array}{cccccccccccccccccccc}0&{\cos \theta }&{\sin \theta }\\{\cos \theta }&{\sin \theta }&0\\{\sin \theta }&0&{\cos \theta }\end{array}} \right|$ is equal to…………….

Since, $\cos 2\theta = 0$

$\Rightarrow$ $\cos 2\theta = \cos \frac{\pi }{2} \Rightarrow 2\theta = \frac{\pi }{2}$

$\Rightarrow$ $\theta = \frac{\pi }{4}$

$\therefore$ $\sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$ and $\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$

${\left| {\begin{array}{cccccccccccccccccccc}0&{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}\end{array}} \right|^2}$

Expanding along ${R_1}$,
$= {\left[ { - \frac{1}{{\sqrt 2 }}\left( {\frac{1}{2}} \right) + \frac{1}{{\sqrt 2 }}\left( { - \frac{1}{2}} \right)} \right]^2} = {\left[ {\frac{{ - 2}}{{2\sqrt 2 }}} \right]^2} = {\left( {\frac{{ - 1}}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}$