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If a plane cuts the coordinate axes at $A, B$ and $C$ respectively such that the centroid of the triangle $A B C$ is $(6,6,3)$, then the equation of that plane.
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Verified Answer
The correct answer is:
$x+y+2 z=18$
Let the equation of plane is
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
Centroid of triangle formed by the coordinates axes at $(a, 0,0),(0, b, 0)$ and $(0,0, c)$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ Given $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(6,6,3)$
$$
\therefore a=18, b=18, c=9
$$
Now, equation of plane is
$$
\begin{aligned}
& \frac{x}{18}+\frac{y}{18}+\frac{z}{9}=1 \\
& x+y+2 z=18
\end{aligned}
$$
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
Centroid of triangle formed by the coordinates axes at $(a, 0,0),(0, b, 0)$ and $(0,0, c)$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ Given $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(6,6,3)$
$$
\therefore a=18, b=18, c=9
$$
Now, equation of plane is
$$
\begin{aligned}
& \frac{x}{18}+\frac{y}{18}+\frac{z}{9}=1 \\
& x+y+2 z=18
\end{aligned}
$$
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