$\left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right| = $

If $A$ is a square matrix of order $n$ and $A = kB$,where $k$ is a scalar,then $|A|=$

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

If $\theta \in \left(0, \frac{\pi}{2}\right)$,then $\left|\begin{array}{ccc} (\sin \theta+\operatorname{cosec} \theta)^2 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \\ (\cos \theta+\sec \theta)^2 & (\cos \theta-\sec \theta)^2 & 2020 \\ (\tan \theta+\cot \theta)^2 & (\tan \theta-\cot \theta)^2 & 2020 \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 $2^{a_1}, 2^{a_2}, 2^{a_3}, \dots, 2^{a_n}$ are in $G.P.$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_{n+1} & a_{n+2} & a_{n+3} \\ a_{2n+1} & a_{2n+2} & a_{2n+3} \end{array} \right|$ is equal to