If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$

If $a,b,c$ are distinct and rational numbers then $\left| {\begin{array}{*{20}{c}}
{\left( {{a^2} + {b^2} + {c^2}} \right)}&{ab + bc + ca}&{ab + bc + ca}\\
{ab + bc + ca}&{\left( {{a^2} + {b^2} + {c^2}} \right)}&{\left( {bc + ca + ab} \right)}\\
{ab + bc + ca}&{\left( {ab + bc + ca} \right)}&{\left( {{a^2} + {b^2} + {c^2}} \right)}
\end{array}} \right|$ is always 

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 $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$