A) \[\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\]
B) \[\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=1\]
C) \[\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=3\]
D) None of these
Correct Answer: A
Solution :
Let the co-ordinates of A, B and C be (a,0,0), (0,b,0) and (0,0,c) respectively. The equation of the plane is \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] Also, it passes through (p, q, r). So,\[\frac{p}{a}+\frac{q}{b}+\frac{z}{c}=1\] Also equation of sphere passes through A, B, C will be \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-ax-by-cz=0\] If its centre \[({{x}_{1}},{{y}_{1}},{{z}_{1}})\], then \[{{x}_{1}}=\frac{a}{2},{{y}_{1}}=\frac{b}{2},{{z}_{1}}=\frac{c}{2}\] \ \[a=2{{x}_{1}},\,\,b=2{{y}_{1}},\,\,c=2{{z}_{1}}\] \ Locus of centre of sphere \[\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2\].
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