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If the plane $3 x-2 y-z-18=0$ meets the coordinate axes in $A, B, C$ then the centroid of $\triangle A B C$ is
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Verified Answer
The correct answer is:
$(2,-3,-6)$
Given equation $3 x-2 y-z=18$ can be rewritten as
$$
\begin{aligned}
\frac{x}{18 / 3}-\frac{y}{18 / 2}-\frac{z}{18} & =1 \\
\frac{x}{6}-\frac{y}{9}-\frac{z}{18} & =1
\end{aligned}
$$
$\therefore$ Points of a coordinates axes are $A(6,0,0), B(0,-9,0)$ and $C(0,0,-18)$.
$\therefore$ Centroid of a triangle
$$
\begin{aligned}
& =\left(\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}\right) \\
& =(2,-3,-6)
\end{aligned}
$$
$$
\begin{aligned}
\frac{x}{18 / 3}-\frac{y}{18 / 2}-\frac{z}{18} & =1 \\
\frac{x}{6}-\frac{y}{9}-\frac{z}{18} & =1
\end{aligned}
$$
$\therefore$ Points of a coordinates axes are $A(6,0,0), B(0,-9,0)$ and $C(0,0,-18)$.
$\therefore$ Centroid of a triangle
$$
\begin{aligned}
& =\left(\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}\right) \\
& =(2,-3,-6)
\end{aligned}
$$
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