If  $\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{2n}\\ 1& x^2& p\\ x^{n+5}& x^{p+6}& x^{2n+5}\end{array}\right|=0$, then  $p$ is given by

$x^n(x^{2n+7}-p{x}^{p+6})-(x^{3x+7}-x^{p+6+2n})+x^{n+5}(p{x}^{n+2}-x^{2n+2})=0$

$\Rightarrow x^{3n+7}-p{x}^{p+n+6}-x^{3n+7}+x^{p+2n+6}+p{x}^{2n+7}-x^{3n+7}=0$

$\Rightarrow p(x^{2n+7}-x^{p+n+6})+(x^{p+2n+6}-x^{3n+7})=0$

To get cancelled,

$2n+7=p+n+6$

$p+2n+6=3n+7$

$p=n+1$

If $p=x^n$, then two rows are equal.

So the determinant is 0.