Question

A variable plane at a distance of $1$ unit from the origin cuts the coordinates axes at $A,B$ and $C.$ If the centroid $D(x,y,z)$ 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 1

The correct option is D $9$

Let the equation of the variable plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,$ which meets the axes at $A(a,0,0),B(0,b,0)$ and $C(0,0,c).$
The centroid of $△ABCis(\frac{a}{3},\frac{b}{3},\frac{c}{3})$ and it satisfies the relation $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$
$\Rightarrow \frac{9}{a^2}+\frac{9}{b^2}+\frac{9}{c^2}=k...\left(1\right)$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{k}{9}$

Also, it is given that the distance of the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ from $(0,0,0)$ is $1$ unit. Therefore,
$\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=1$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get $\frac{k}{9}=1\Rightarrow k=9$