A) \[x+y+z=6\]
B) \[\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1\]
C) \[\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\]
D) \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}\]
Correct Answer: C
Solution :
Let \[a,b,c\] be the intercepts of the variable plane on the \[x,y,z\] axes respectively, then the equation of the plane is \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]. And the point of intersection of the \[xy,yz,zx\]planes parallel to the planes is clearly the point\[(a,b,c)\]. Since the point \[(3,2,1)\] lies on the variable plane, we have\[\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\]. Thus the required locus is\[\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\] So the answer is option C.
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