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If a triangle $A B C$ with two vertices $A(5,4,6)$ and $B(1,-1,3)$ has its centroid at $\left(\frac{10}{3}, 2, \frac{11}{3}\right)$, then the third vertex $C$ is
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Verified Answer
The correct answer is:
$(4,3,2)$
We have, $\triangle A B C$
$$
A=(5,4,6), B=(1,-1,3), C=(x, y, z)
$$
Centroid of $\triangle A B C$ is $\left(\frac{10}{3}, 2, \frac{11}{3}\right)$
$$
\begin{aligned}
\therefore \quad \frac{10}{3} & =\frac{5+1+x}{3} \Rightarrow x=4 \\
2 & =\frac{4-1+y}{3} \Rightarrow y=3 \\
\frac{11}{3} & =\frac{6+3+z}{3} \Rightarrow z=2
\end{aligned}
$$
$\therefore$ Vertices $C \equiv(4,3,2)$
$$
A=(5,4,6), B=(1,-1,3), C=(x, y, z)
$$
Centroid of $\triangle A B C$ is $\left(\frac{10}{3}, 2, \frac{11}{3}\right)$
$$
\begin{aligned}
\therefore \quad \frac{10}{3} & =\frac{5+1+x}{3} \Rightarrow x=4 \\
2 & =\frac{4-1+y}{3} \Rightarrow y=3 \\
\frac{11}{3} & =\frac{6+3+z}{3} \Rightarrow z=2
\end{aligned}
$$
$\therefore$ Vertices $C \equiv(4,3,2)$
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