Question

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

Answer: (D)

$9$

Equation of plane at a distance $1$ from origin is given by,
$lx+my+nz=1$, where $l,m,n$ are direction cosine of normal to the plane along distance
Thus intercept on the axes are,
$A=(\frac{1}{l},0,0),B=(0,\frac{1}{m},0),C=(0,0,\frac{1}{n})$
So centroid of ABC is,
$x=\frac{1}{3l},y=\frac{1}{3m},z=\frac{1}{3n}$
Also we know,
$l^2+m^2+n^2=1$
$\Rightarrow \frac{1}{9x^2}+\frac{1}{9y^2}+\frac{1}{9z^2}=1$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=9$
Hence, required value of $k$ is $9$.