A) 9
B) 3
C) \[\frac{1}{9}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
A variable plane \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] cuts the coordinate axes at A(a, 0, 0), B (0, b, 0), C (0, 0, c) and its distance from origin is 1. \[\therefore \]\[\frac{1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}}}=1\]\[\Rightarrow \]\[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=1\] ?(i) Coordinates of centroid \[p(x,y,z)=\left( \frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3} \right)\] \[\Rightarrow \]\[x=\frac{a}{3},y=\frac{b}{3},z=\frac{c}{3}\] ..(ii) From Eqs. (i) and (ii), we get \[\frac{1}{9{{x}^{2}}}+\frac{1}{9{{y}^{2}}}+\frac{1}{9{{z}^{2}}}=1\]\[\Rightarrow \]\[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=9=(k)\]given\[\Rightarrow \] k = 9
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