If $a, b, c$  are all different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} – 1}\\b&{{b^3}}&{{b^4} – 1}\\c&{{c^3}}&{{c^4} – 1}\end{array}\,} \right|$ = $0$ , then the value of $abc(ab + bc + ca)$ is

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to

If the determinant $\left| {\begin{array}{*{20}{c}}{a\, + \,p}&{1\, + \,x}&{u\, + \,f}\\ {b\, + \,q}&{m\, + \,y}&{v\, + \,g}\\{c\, + \,r}&{n\, + \,z}&{w\, + \,h}\end{array}} \right|$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$, is

Let $z=\frac{-1+\sqrt{3} i}{2}$, where $i=\sqrt{-1}$, and $r, s \in\{1,2,3\}$. Let$P=\left[\begin{array}{cc}(-z)^r & z^{2 s} \\ z^{2 s} & z^r\end{array}\right]$ and $I$ be the identity matrix of order $2$ . Then the total number of ordered pairs $(r, s)$ for which $P^2=-I$ is

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi), \operatorname{are}$