Question

A plane intersects the $x,y,z$ axes at $A(a,0,0)$ , $B(0,b,0)$ , $C(0,0,c)$ respectively. lf $(p,q,r)$ is the centroid of the $\Delta ABC$ and the equation of the plane is $ax+by+cz-d=0$ then $a+b+c+d=$

• A. $p+q+r$
• B. $p+q+r+2$
• C. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-3$
• D. $p+q+r+3$

Answer 1

Answer: (C)

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-3$

$A$ plane intersects the $x,y,z$ axes at $A(a,0,0),B(0,b,0),C(0,0,c)$ respectively.
Centroid of $ABC$ is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$.
Therefore, $p=\frac{a}{3}$, $q=\frac{b}{3}$ and $r=\frac{c}{3}$, equation of plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow \frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1$
Therefore, $a+b+c+d=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-3$