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A plane cuts the coordinate axes $X, Y, Z$ at $A$, $B, C$ respectively such that the centroid of the $\triangle A B C$ is $(6,6,3)$. Then the equation of that plane is
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Verified Answer
The correct answer is:
$x+y+2 z-18=0$
Let the equation of the plane is
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
By the definition of the centroi $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(6,6,3)$
Therefore,
$$
\begin{aligned}
a & =18 \\
b & =18 \\
c & =9
\end{aligned}
$$
The equation of the plane becomes
$$
\begin{array}{rlrl}
& & \frac{x}{18}+\frac{y}{18}+\frac{z}{9} & =1 \Rightarrow \frac{x+y+2 z}{18}=1 \\
\Rightarrow \quad & x+y+2 z & =18 \\
\Rightarrow \quad x+y+2 z-18 & =0
\end{array}
$$
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
By the definition of the centroi $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(6,6,3)$
Therefore,
$$
\begin{aligned}
a & =18 \\
b & =18 \\
c & =9
\end{aligned}
$$
The equation of the plane becomes
$$
\begin{array}{rlrl}
& & \frac{x}{18}+\frac{y}{18}+\frac{z}{9} & =1 \Rightarrow \frac{x+y+2 z}{18}=1 \\
\Rightarrow \quad & x+y+2 z & =18 \\
\Rightarrow \quad x+y+2 z-18 & =0
\end{array}
$$
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