Q75 / 155 · WBJEE 2024 · Mathematics

If $\left|\begin{array}{lll}x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3}\end{array}\right|=(x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$, then

  • A
    $\mathrm{k}=-3$
  • B
    k=3
  • C
    k=1
  • Correct answer
    D
    k=-1

Solution

$\begin{aligned} \text { Hint : } & (x y z)^k\left|\begin{array}{ccc}1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3\end{array}\right| \\ & =(x y z)^k(x-y)(z-x)-(y-z)(x y+y z+z x) \\ & =(x y z)^{k+1}(x-y)(z-x)-(y-z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\ & k+1=0 \\ & k=-1\end{aligned}$

Hence, the correct answer is option: 'D'.

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