If begin{vmatrix} x^k & x^{k+2} & x^{k+3} y^ | vedclass

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(D) Given the determinant $\Delta = \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \ y^k & y^{k+2} & y^{k+3} \ z^k & z^{k+2} & z^{k+3} \end{vmatrix}$.Takin

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$k=-1$

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(D) Given the determinant $\Delta = \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \ y^k & y^{k+2} & y^{k+3} \ z^k & z^{k+2} & z^{k+3} \end{vmatrix}$.Taking $x^k, y^k, z^k$ common from $R_1, R_2, R_3$ respectively:$\Delta = (xyz)^k \begin{vmatrix} 1 & x^2 & x^3 \ 1 & y^2 & y^3 \ 1 & z^2 & z^3 \end{vmatrix}$.The value of the determinant $\begin{vmatrix} 1 & x^2 & x^3 \ 1 & y^2 & y^3 \ 1 & z^2 & z^3 \end{vmatrix}$ is $(x-y)(y-z)(z-x)(xy+yz+zx)$.So, $\Delta = (xyz)^k (x-y)(y-z)(z-x)(xy+yz+zx)$.We can write $(xy+yz+zx) = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}ight)$.Therefore, $\Delta = (xyz)^k (xyz) (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}ight) = (xyz)^{k+1} (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}ight)$.Comparing this with the given expression $(x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}ight)$, we must have $(xyz)^{k+1} = 1$.This implies $k+1 = 0$, so $k = -1$.

If $\begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} = (x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$, then find the value of $k$.

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