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If a plane meets the coordinate axes at $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in such a way that the centroid of $\triangle \mathrm{ABC}$ is at the point $(1,2,3)$, then equation of the plane is
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Verified Answer
The correct answer is:
$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
Let the equation of the required plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
This meets the coordinate axes at $A, B$ and $\mathrm{C}$, the coordinates of the centroid of $\triangle \mathrm{ABC}$
are $\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)$
$$
\begin{array}{l}
\therefore \frac{\mathrm{a}}{3}=1, \frac{\mathrm{b}}{3}=2, \frac{\mathrm{c}}{3}=3 \\
\Rightarrow \mathrm{a}=3, \mathrm{~b}=6, \mathrm{c}=9
\end{array}
$$
Hence, the equation of the plane is
$$
\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1
$$
This meets the coordinate axes at $A, B$ and $\mathrm{C}$, the coordinates of the centroid of $\triangle \mathrm{ABC}$
are $\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)$
$$
\begin{array}{l}
\therefore \frac{\mathrm{a}}{3}=1, \frac{\mathrm{b}}{3}=2, \frac{\mathrm{c}}{3}=3 \\
\Rightarrow \mathrm{a}=3, \mathrm{~b}=6, \mathrm{c}=9
\end{array}
$$
Hence, the equation of the plane is
$$
\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1
$$
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