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If a plane meets the coordinate axes at $A, B$ and $C$ such that the centroid of the triangle is $(1,2,4)$, then the equation of the plane is
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Verified Answer
The correct answer is:
$4 x+2 y+z=12$
Let the equation of the plane is
$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$
Then, $\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)$ and $\mathrm{C}(0,0, \gamma)$ are the points on the coordinate axes, The centroid of the triangle is $(1,2,4)$.
$\begin{array}{l}
\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3 \\
\frac{\beta}{3}=2 \Rightarrow \beta=6
\end{array}$
and $\frac{\gamma}{3}=4 \Rightarrow \gamma=12$
$\therefore \quad$ The equation of the plane is
$\begin{aligned}
& \frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \\
\Rightarrow & 4 x+2 y+z=12
\end{aligned}$
$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$
Then, $\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)$ and $\mathrm{C}(0,0, \gamma)$ are the points on the coordinate axes, The centroid of the triangle is $(1,2,4)$.
$\begin{array}{l}
\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3 \\
\frac{\beta}{3}=2 \Rightarrow \beta=6
\end{array}$
and $\frac{\gamma}{3}=4 \Rightarrow \gamma=12$
$\therefore \quad$ The equation of the plane is
$\begin{aligned}
& \frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \\
\Rightarrow & 4 x+2 y+z=12
\end{aligned}$
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