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

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

If $D =\left| {\begin{array}{*{20}{c}}{{a^2}\, + \,\,1}&{ab}&{ac}\\{ba}&{{b^2}\, +\,\,1}&{bc}\\{ca}&{cb}&{{c^2}\, + \,\,1}\end{array}} \right|$ then $D =$

Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$