The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

If $\mathrm{a, b, c},$ are in $\mathrm{A.P}$, then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is

If $\omega $is a cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\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

$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$