If $\begin{vmatrix}x^n & x^{n+2} & x^{n+3}\\y^n & y^{n+2} & y^{n+3}\\ z^n & z^{n+2} & z^{n+3} \end{vmatrix} $

$=(x-y) (y-z) (z-x) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right),$ then  n equals

-1

The correct answer is option (3) : -1

The degree of the determinant is $n+(n+2) + (n+3)= 3n+ 5.$ and the degree of the expression on RHS is 2.

$∴3n+ 5= 2 ⇒n=-1$