If $\begin{vmatrix}x^n&x^{n+2}&x^{n+3}\\y^n&
y^{n+2}& y^{n+3}\\z^n&
z^{n+2}& z^{n+3}\end{vmatrix}=(x-y) (y-z) (z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$, then n equals
Answer & explanation
Correct answer: option 2
The degree of the determinant is $n+(n+2)+(n+3)=3n+5$ and the degree of the expression on RHS is 2.
$∴3n+5=2⇒n=-1$