A) \[x+y+z=1\]
B) \[x+y+z=2\]
C) \[\frac{x}{3}+\frac{y}{3}+\frac{z}{3}=3\]
D) \[x+y+z=\frac{1}{3}\]
Correct Answer: A
Solution :
Let the equation of plane be \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] ? (i) Since, the plane meet the coordinate axes at A B and C. \[\therefore \]The coordinates of these points are \[A(a,0,0),B(0,b,0)\] and \[C(0,0,c),\] respectively. Then, centroid of \[\Delta ABC\]is\[\left( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right),\] but it is given \[\left( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right)\] \[\therefore \] \[\frac{a}{3}=\frac{1}{3},\frac{b}{3}=\frac{1}{3},\frac{c}{3}=\frac{1}{3}\] \[\Rightarrow \] \[a=1,b=1,c=1\] On putting the value of a, b and c in Eq. (i). we get \[\frac{x}{y}+\frac{y}{1}+\frac{z}{1}=1\Rightarrow x+y+z=1\]
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