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 centroid of triangle ABC is $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{k}{p^2}$ where k equals

• A. 3
• B. 1
• C. 4
• D. 2

Answer 1

The correct option is B 1

General equation of plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Perpendicular distance from origin
$3p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$ …(i)
Locus of centroid of triangle $(h,k,m)$
$h=\frac{a}{3},k=\frac{b}{3},m=\frac{c}{3}$ …(ii)
Solve (i) and (ii)
$3p=\frac{1}{\sqrt{\frac{1}{{\left(3h\right)}^2}+\frac{1}{{\left(3k\right)}^2}+\frac{1}{{\left(3m\right)}^2}}}$
$\Rightarrow \frac{1}{9p^2}=\frac{1}{9h^2}+\frac{1}{9k^2}+\frac{1}{9m^2}$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}\Rightarrow k=1$