$\left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \\ 1 & ca+bd & c^2a^2+b^2d^2 \\ 1 & ab+cd & a^2b^2+c^2d^2\end{array}\right|=$

If $\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$ and $0 < \theta < \frac{\pi}{2}$,then $\cos 4 \theta$ is equal to

If $x, y, z \in \mathbb{R}$,then the value of the determinant $\left|\begin{array}{lll}\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|$ is:

By using properties of determinants,show that:
$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a)$

$A$ value of $\theta$ lying between $0$ and $\frac{\pi}{2}$ and satisfying $\left|\begin{array}{ccc} 1+\sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1+\cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1+4\sin 4\theta \end{array}\right| = 0$ is:

If $a \neq b \neq c$, $\Delta_1=\left|\begin{array}{lll}1 & a^2 & b c \\ 1 & b^2 & c a \\ 1 & c^2 & a b\end{array}\right|$, $\Delta_2=\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ and $\frac{\Delta_1}{\Delta_2}=\frac{6}{11}$, then $11(a+b+c)=$