$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

The value of the determinant $\left|\begin{array}{ccc}a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b\end{array}\right|$ is

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

If $A$ and $B$ are two square matrices with $\det(A) = 5$ and $\det(B^T \cdot A^T) = -15$,then $\det(B)$ is equal to

$\left| {\begin{array}{*{20}{c}} 1 & 5 & \pi \\ {{\log }_e}e & 5 & {\sqrt 5 } \\ {{\log }_{10}}10 & 5 & e \end{array}} \right| = $

By using properties of determinants,show that:
$\left|\begin{array}{ccc}-a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2}\end{array}\right|=4a^{2}b^{2}c^{2}$