Question

A variable plane which remains at a constant distance 3p from the origin cuts the co-ordinate axes at A, B, C. The locus of the centroid of triangle ABC is

• A. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3p}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{9p^2}$
• C. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Answer 1

The correct option is D

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Let the equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$=1.
Then, A=(a,0,0), B=(0,b,0), C=(0,0,c). The centroid of ABC is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$
Distance of the plane from the origin.
$3p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{9p^2}$
Let (x,y,z) be the coordinates of the centroid. Then,
$x=\frac{a}{3}\Rightarrow a=3x.Also,b=3y,c=3z\Rightarrow \frac{1}{9x^2}+\frac{1}{9y^2}+\frac{1}{9z^2}=\frac{1}{9p^2}\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$