# A variable plane which remains at a constant distance 3p from the origin cuts the co-ordinate axes at A, B & C. Show that the locus of the centroid of the $\bigtriangleup ABC\;is\; \Large\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Let the variable plane at distance 3p from the origin be :
ax + by + cz = 3p - - - - - eq(1)
where
a, b, and c are the direction-cosines of the normal to the plane

This meet the x-axis in A. Thus, y = z = 0, substituting to (1),
ax = 3p,
x = 3p/a. Hence,
coordinate A = (3p/a, 0, 0).
Similarly, coordinate B and C are :
B = (0, 3p/b, 0),
C = (0, 0, 3p/c)

Now, let the centroid of ABC is G(x, y, z).
Using the formula of centroid :
(x, y, z) = {(x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3},
G(x, y, z) = {(3p/a + 0 + 0)/3 , (0 + 3p/b + 0)/3, (0 + 0 + 3p/c)/3},
G(x, y, z) = (p/a, p/b, p/c)

Since a, b, and c are the d.c’s
then a² + b² + c² = 1.
But a = p/x, b = p/y, and c = p/z. So :
p²/x² + p²/y² + p²/z² = 1

1/x² +1/y² +1/z² =p²

hope u find this heplful