# 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 $\bigtriangleup ABC\;is\; \Large\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$

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)
O=(0,0,0)
now the centroid of the tedtrahedron is given by the formula:-
{(x1+x2+x3+x4)/4,(y1+y2+y3+y4)/4,(z1+z2+z3+z4)/4}
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