Question

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

• A. $\sqrt{14}$
• B. $14$
• C. $\sqrt{13}$
• D. $0$

Answer 1

The correct option is A $\sqrt{14}$

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