If $\left| {\begin{array}{*{20}{c}}   {a - b}&{b - c}&{c - a} \\    {b - c}&{c - a}&{a - b} \\    {c - a + 1}&{a - b}&{b - c}  \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$ is not depend

If $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :

Using properties of determinants, prove this:

$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

If $a, b, c$  are all different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} - 1}\\b&{{b^3}}&{{b^4} - 1}\\c&{{c^3}}&{{c^4} - 1}\end{array}\,} \right|$ = $0$ , then the value of $abc(ab + bc + ca)$ is

$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $