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

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x - 1)}&{(x + 1)x}\\{3x(x - 1)}&{x(x - 1)(x - 2)}&{(x + 1)x(x - 1)}\end{array}} \right|$ then $f(100)$ is equal to

If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ - x}}} \right)}^2}}&{{{\left( {{a^x}\, - \,\,{a^{ - x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ - y}}} \right)}^2}}&{{{\left( {{b^y}\, - \,\,{b^{ - y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ - z}}} \right)}^2}}&{{{\left( {{c^z}\, - \,\,{c^{ - z}}} \right)}^2}}&1\end{array}} \right|$ $=$

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 $1+x y z=0$.

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=$ $Ax ^{3}+ Bx ^{2}+ Cx + D ,$ then $B + C$ is equal to