If centroid of the tetrahedron OABC, where A,B,C are given by (a, 2, 3),(1, b, 2) and (2, 1, c) respectively be (1, 2, -1), then distance of P(a,b,c) from origin is equal to

$\sqrt{107}$

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$\sqrt{14}$

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$\sqrt{\frac{107}{14}}$

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$N o n e o f t h e s e$

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Solution

The correct option is A
$\sqrt{107}$

Centroid $\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)$
$\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}$ .

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If centroid of the tetrahedron OABC, where A,B,C are given by (a, 2, 3),(1, b, 2) and (2, 1, c) respectively be (1, 2, -1), then distance of P(a,b,c) from origin is equal to

The correct option is A √107 Centroid ≡(∑x4,∑y4,∑z4)=(1,2,−1) ⇒a=1,b=5,c=−9;∴√a2+b2+c2=√107.

The correct option is A
$\sqrt{107}$

Centroid $\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)$
$\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}$ .