$\left|\begin{array}{ccc}0 & \sin 2 \alpha & -\cos ^2 \alpha \\ -\sin ^2 \alpha & 0 & \sin \alpha \sin \beta \\ -\cos \alpha \sin \beta & 2 \sin ^2 \beta & 0\end{array}\right|$
Answer & explanation
Correct answer: option 1
$=0(0×0-2sin^2β×sinα\, sinβ)-sin2α(-sin^2α×0-(-cosα\, sinβ)(sinα\, sinβ))+-cos^2α(-sin^2α×2sin^2β)-0×(-cosα\, sinβ)$
$=-sin^2α(sinα\, cosα\, sin^2β)-cos^2α(-2sin^2α\, sin^2β)$
$=-2sin^2α\, cos^2α\, sin^2β+2sin^2α\, cos^2α\, sin^2β$
$= 0$