$A (2, 3, - 4)$ , $B (- 3, 3, - 2)$ , $C (- 1, 4, 2)$ and $D (3, 5, 1)$ are the vertices of a tetrahedron. If E, F, G are the centroid of its faces containing the point A, then the centroid of the triangle EFG is

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$(\frac{- 1}{9}, \frac{12}{3}, \frac{1}{9})$

$(\frac{4}{9}, \frac{11}{3}, \frac{- 10}{9})$

$(\frac{1}{4}, \frac{15}{4}, \frac{- 3}{4})$

$(\frac{1}{9}, \frac{15}{9}, \frac{- 3}{9})$

answer is C.

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Detailed Solution

We have, $A (2, 3, - 4), B (- 3, 3, - 2), C (- 1, 4, 2), D (3, 5, 1)$
Now, E = centroid of $Δ A B C$
$= (\frac{2 - 3 - 1}{3}, \frac{3 + 3 + 4}{3}, \frac{- 4 - 2 + 2}{3})$
$= (\frac{- 2}{3}, \frac{10}{3}, \frac{- 4}{3})$
F = centroid of $Δ A B D$
$= (\frac{2 - 3 + 3}{3}, \frac{3 + 3 + 5}{3}, \frac{- 4 - 2 + 1}{3}) = (\frac{2}{3}, \frac{11}{3}, \frac{- 5}{3})$
G = centroid of $Δ A C D$
$= (\frac{2 - 1 + 3}{3}, \frac{3 + 4 + 5}{3}, \frac{- 4 + 2 + 1}{3}) = (\frac{4}{3}, 4, \frac{- 1}{3})$
Let centroid of $Δ E F G$ is H
$∴ H = (\frac{\frac{- 2}{3} + \frac{2}{3} + \frac{4}{3}}{3}, \frac{\frac{10}{3} + \frac{11}{3} + 4}{3}, \frac{\frac{- 4}{3} \frac{- 5}{3} \frac{- 1}{3}}{3})$
$= (\frac{4}{9}, \frac{11}{3}, \frac{- 10}{9})$