Question

A plane meet the co-ordinate axes in $A,B,C$ such that the centroid of triangle $ABC$ is $(a,b,c)$. If equation of plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k(k\ne 0)$, then the value of $k$ equals

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

Given equation of plane is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k$
$\frac{x}{ak}+\frac{y}{bk}+\frac{z}{ck}=1$
So, the plane intersects $x$-axis at $(ak,0,0)$, y-axis at $(0,bk,0)$ and $z$-axis at $(0,0,ck)$
Coordinates of centroid of triangle $ABC$ is $(\frac{ak}{3},\frac{bk}{3},\frac{ck}{3})$
Given coordinates of centroid are $(a,b,c)$
$\Rightarrow \frac{ak}{3}=a,\frac{bk}{3}=b,\frac{ck}{3}=c$
$\Rightarrow k=3$