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A plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle A B C$ is $(2,3,5)$. Then, the equation of that plane is
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Verified Answer
The correct answer is:
$15 x+10 y+6 z=90$
Let the equation of plane is, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ where $a, b, c$ are $x, y$ and $z$ intercepts
$\therefore$ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.
$\begin{array}{rlrl}
\text { Given, }\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) & =(2,3,5) \\
\therefore \quad & a=6, b & =9, c=15
\end{array}$
Equation of plane is, $\frac{x}{6}+\frac{y}{9}+\frac{z}{15}=1$
$15 x+10 y+6 z=90$
$\therefore$ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.
$\begin{array}{rlrl}
\text { Given, }\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) & =(2,3,5) \\
\therefore \quad & a=6, b & =9, c=15
\end{array}$
Equation of plane is, $\frac{x}{6}+\frac{y}{9}+\frac{z}{15}=1$
$15 x+10 y+6 z=90$
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