The circum radius of the triangle formed by the points $(1, 2, - 3), (2, - 3, 1)$ and $(- 3, 1, 2)$ is:

\sqrt{14}

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14

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\sqrt{13}

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0

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Solution

The correct option is A $\sqrt{14}$
Let $A (1, 2, - 3), B (2, - 3, 1), C (- 3, 1, 2)$
$| A B | = \sqrt{1^{2} + 5^{2} + 4^{2}} = \sqrt{42}$
$| B C | = \sqrt{5^{2} + 4^{2} + 1^{2}} = \sqrt{42}$
$| A C | = \sqrt{4^{2} + 1^{2} + 5^{2}} = \sqrt{42}$
It is an equilateral triangle. Hence, the value of $R$ will be $\frac{a}{\sqrt{3}}$ $= \sqrt{14}$

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The circum radius of the triangle formed by the points (1,2,−3),(2,−3,1) and (−3,1,2) is:

The correct option is A √14Let A(1,2,−3),B(2,−3,1),C(−3,1,2)|AB|=√12+52+42=√42|BC|=√52+42+12=√42|AC|=√42+12+52=√42It is an equilateral triangle. Hence, the value of R will be a√3 =√14

The circum radius of the triangle formed by the points $(1, 2, - 3), (2, - 3, 1)$ and $(- 3, 1, 2)$ is: