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If the centroid of the tetrahedron $A B C D$ is $G (0, - 1, \frac{5}{2})$ and three vertices are $A (- 2, 3, 4), B (3, - 4, 2), C (2, - 5, 1)$ then the distance of centroid $G$ from fourth vertex $D$ is

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$\frac{\sqrt{37}}{2}$

$\frac{\sqrt{73}}{2}$

$\frac{\sqrt{73}}{4}$

$\frac{\sqrt{37}}{4}$

answer is C.

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Detailed Solution

$G = (0, - 1, \frac{5}{2}) A = (- 2, 3, 4) B = (3, - 4, 2) C = (2, - 5, 1) D = 4 G - A - B - C = (- 3, 2, 3) G D = \frac{\sqrt{73}}{2}$

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