If (2,3,4) is the centroid of the tetrahedron for which  (2,3,−1),(3,0,−2),(−1,4,3)are three vertices , then the distance of the fourth vertex from the origin is

# If $\left(2,3,4\right)$ is the centroid of the tetrahedron for which  $\left(2,3,-1\right),\left(3,0,-2\right),\left(-1,4,3\right)$are three vertices , then the distance of the fourth vertex from the origin is

1. A

$\sqrt{33}$

2. B

$2\sqrt{33}$

3. C

$3\sqrt{33}$

4. D

$4\sqrt{33}$

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### Solution:

The centroid of the tetrahedron formed by the vertices  $\left({x}_{1},{y}_{1},{z}_{1}\right),\left({x}_{2},{y}_{2},{z}_{2}\right),\left({x}_{3},{y}_{3},{z}_{3}\right),\left({x}_{4},{y}_{4},{z}_{4}\right)$ is $G\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}}{4},\frac{{y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}}{4},\frac{{z}_{1}+{z}_{2}+{z}_{3}+{z}_{4}}{4}\right)$

Hence, Fourth vertex is

$D=4G-A-B-C=\left(8-2-3+1,12-3-4,16+1+2-3\right)=\left(4,5,16\right)$

The distance between the fourth vertex and the centroid is $DG=\sqrt{16+25+256}=\sqrt{297}$

Therefore, the required distance is $\overline{)3\sqrt{33}}$  +91

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