A) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=1\]
B) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=3\]
C) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=9\]
D) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=\frac{1}{9}\]
Correct Answer: A
Solution :
Let Centroid be \[(h,\,k,\,l)\] \[\therefore \,\,\,x-\,\text{intp}\,\,\text{=3h}\,\,\,\,\,\text{Y-intp}\,\text{=3k,}\,\,\text{3-int}\,\,\text{= 3l}\] Equ. \[\frac{x}{3h}+\frac{y}{3k}+\frac{z}{3\ell }=1\] dist from \[(0,0,0)\] \[\left| \frac{-1}{\sqrt{\frac{1}{9{{h}^{2}}}+\frac{1}{9{{k}^{2}}}+\frac{1}{9{{l}^{2}}}}} \right|=3\] \[=1\]
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