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A variable plane is at constant distance p from the origin and meet the axes in A, B & C. Show that the locus of the centroid of the tetrahedron $\bigtriangleup ABC\;is\; \Large\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$

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i guess there is a small correction in the question its "tetrahedron OABC"
let lx+my+nz+d=p be the eq of the plane where p is the distance from the origin and l,m,n are the dc's
since the plane meets the co ordinate axes 
This meet the x-axis in A. Thus, y = z = 0, substituting to (1), 
lx = p, 
x = p/l. Hence,  
coordinate A = (p/l, 0, 0). 
Similarly, coordinate B and C are : 
B = (0, p/m, 0), 
C = (0, 0, p/n)
 now the centroid of the tedtrahedron is given by the formula:-
G(x, y, z) = {(p/4l + 0 + 0+0)/4 , (0 + p/4m + 0+0)/4, (0 + 0 + p/4n+0)/4}, 
G(x, y, z) = (p/4l, p/4m, p/4n)
Since a, b, and c are the d.c’s 
then a² + b² + c² = 1. 
But a = p/4x, b = p/4y, and c = p/4z. So : 
p²/16x² + p²/16y² + p²/16z² = 1
1/x² +1/y² +1/z² =16/p²


answered Feb 4 by loserlady2lol

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