$\left|\begin{array}{ccc}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b\end{array}\right|$ is equal to

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| = $

If $\Delta = \begin{vmatrix} x+y+z^2 & x^2+y+z & x+y^2+z \\ z^2 & x^2 & y^2 \\ x+y & y+z & x+z \end{vmatrix}$,(where $x \neq y \neq z$ and $x, y, z \in \mathbb{R} - \{0\}$),then $\Delta = $ . . . . . . .

For non-zero $a, b, c$,if $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$,then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

By using properties of determinants,show that:
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

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