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If ∣1+x21111+y21111+z2∣=k(1+1x2+1y2+1z2)\begin{vmatrix} 1+x^2 & 1 & 1 \\ 1 & 1+y^2 & 1 \\ 1 & 1 & 1+z^2 \end{vmatrix} = k\left(1 + \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right) then kk is equal to

Solution

✅ Correct Option: 3

Expanding the determinant gives x2y2z2+x2y2+y2z2+z2x2x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2. Taking x2y2z2x^2y^2z^2 common yields x2y2z2(1+1z2+1x2+1y2)x^2y^2z^2\left(1+\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}\right), so k=x2y2z2k=x^2y^2z^2.

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