Question
A variable plane at a distance of the one unit from the origin cuts the coordinates axes at $$A,\,B$$ and $$C.$$ If the centroid $$D\left( {x,\,y,\,z} \right)$$ of triangle $$ABC$$ satisfies the relation $$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k,$$ then the value of $$k$$ is :
A.
$$3$$
B.
$$1$$
C.
$$\frac{1}{3}$$
D.
$$9$$
Answer :
$$9$$
Solution :
Let the equation of variable plane be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ which meets the axes at $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$ and $$C\left( {0,\,0,\,c} \right).$$
$$\therefore $$ Centroid of $$\Delta ABC$$ is $$\left( {\frac{a}{3},\,\frac{b}{3},\,\frac{c}{3}} \right)$$
and it satisfies the relation
$$\eqalign{
& \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k \cr
& \Rightarrow \frac{9}{{{a^2}}} + \frac{9}{{{b^2}}} + \frac{9}{{{c^2}}} = k \cr
& \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{k}{9}.....(1) \cr} $$
Also given that the distance of plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$
from (0, 0, 0) is 1 unit.
$$\eqalign{
& \Rightarrow \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }} = 1 \cr
& \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 1.....(2) \cr} $$
From (1) and (2), we get $$\frac{k}{9} = 1\,\,\,\,\,i.e.,\,\,k = 9$$