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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
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The correct answer is:
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
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
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
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
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
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