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If the origin is the centroid of the triangle for which $(-2,3,4)$ and $(3,-1,5)$ are two vertices, then the third vertex is
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Verified Answer
The correct answer is:
$(-1,-2,-9)$
Let the vertices of triangle be
$$
A(x, y, z), B(-2,3,4), C(3,-1,5)
$$
Centroid of $\triangle A B C$ is
$$
\begin{aligned}
\left(\frac{x-2+3}{3}, \frac{y+3-1}{3}\right. & \left., \frac{z+4+5}{3}\right) \\
& =\left(\frac{x+1}{3}, \frac{y+2}{3}, \frac{2+9}{3}\right)
\end{aligned}
$$
Given that centroid of $\triangle A B C$ is origin
$$
\begin{aligned}
& \frac{x+1}{3}=0 \Rightarrow x=-1 \\
& \frac{y+2}{3}=0 \Rightarrow y=-2 \\
& \frac{z+9}{3}=0 \Rightarrow z=-9
\end{aligned}
$$
$\therefore$ Third vertex $(-1,-2,-9)$.
$$
A(x, y, z), B(-2,3,4), C(3,-1,5)
$$
Centroid of $\triangle A B C$ is
$$
\begin{aligned}
\left(\frac{x-2+3}{3}, \frac{y+3-1}{3}\right. & \left., \frac{z+4+5}{3}\right) \\
& =\left(\frac{x+1}{3}, \frac{y+2}{3}, \frac{2+9}{3}\right)
\end{aligned}
$$
Given that centroid of $\triangle A B C$ is origin
$$
\begin{aligned}
& \frac{x+1}{3}=0 \Rightarrow x=-1 \\
& \frac{y+2}{3}=0 \Rightarrow y=-2 \\
& \frac{z+9}{3}=0 \Rightarrow z=-9
\end{aligned}
$$
$\therefore$ Third vertex $(-1,-2,-9)$.
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