$\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right| = $

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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

At what value of $x,$ will $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$

If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.