Question
A variable plane at a distance of $$1$$ unit from the origin cuts the coordinate 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 $$C\left( {0,\,0,\,c} \right)$$
The centroid of $$\Delta ABC$$ is $$\left( {\frac{a}{3},\,\frac{b}{3},\,\frac{c}{3}} \right)$$ and it satisfies the relation $$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k.$$
Thus,
$$\eqalign{
& \frac{9}{{{a^2}}} + \frac{9}{{{b^2}}} + \frac{9}{{{c^2}}} = k \cr
& {\text{or, }}\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{k}{9}......\left( {\text{i}} \right) \cr} $$
Also it is given that the distance of the plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ from $$\left( {0,\,0,\,0} \right)$$ is $$1$$ unit. Therefore,
$$\eqalign{
& \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }} = 1 \cr
& {\text{or, }}\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 1......\left( {{\text{ii}}} \right) \cr} $$
From $$\left( {\text{i}} \right)$$ and $$\left( {\text{ii}} \right)$$, we get $$\frac{k}{9} = 1,\,{\text{i}}{\text{.e}}{\text{., }}k = 9$$