Question

A plane $P$ meets the coordinate axes at $A$, $B$ and $C$ respectively. The centroid of $△ABC$ is given to be $(1,1,2)$. Then the eqaution of the line through this centroid and perpendicular to the plane $P$ is

• A. $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$
  
• B. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-2}{2}$
  
• C. $\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$
  
• D. $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-2}{1}$
  

Answer 1

The correct option is C $\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$


Let $A(\alpha ,0,0),B(0,\beta ,0)$ and $(0,0,\gamma ).$


Then, $G(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})=(1,1,2)$
$\Rightarrow \alpha =3,\beta =3,\gamma =6$

Now, equation of plane is
$\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$
$\Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1$
$\Rightarrow 2x+2y+z=6$
$\therefore$ Required equation of line
$\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$