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 $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of

$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$

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

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.