Q63 / 180 · KCET 2021 · Mathematics

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of $\triangle A B C$ is at the point $(1,2,3)$, then the equation of the plane is

  • A
    $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1$
  • Correct answer
    B
    $\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
  • C
    $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=\frac{1}{3}$
  • D
    $\frac{x}{1}-\frac{y}{2}+\frac{z}{3}=-1$

Solution

Given, plane meets the coordinate axes at $A, B$ and $C$. Let the plane meets the coordinate axes at the points $A(a, 0,0), B(0, b, 0), C(0,0, c)$ The equation of plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...(i)$ Centroid formula says, $\begin{aligned} &x=\frac{x_{1}+x_{2}+x_{3}}{3}, y=\frac{y_{1}+y_{2}+y_{3}}{3} \\ &z=\frac{z_{1}+z_{2}+z_{3}}{3} \end{aligned}$ Centroid of $\triangle A B C=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ Given centroid $=(1,2,3)$ $\Rightarrow\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ Therefore, $a=3, b=6$ and $c=9$ Put the value of $a, b$ and $c$ in equation of plane (i), we get $\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$

Hence, the correct answer is option: 'B'.

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