The circumradius of $∆\mathrm{ABC}$ with vertices A(2, –1, 1), B(1, –3, –5), C(3, –4, –4) is

The given vertices are $A(2, –1, 1), B(1, –3, –5), C(3, –4, –4)$

Lengths of the sides are 

$$\begin{gathered} a=BC=\sqrt{4+1+1}=\sqrt{6} \\ b=CA=\sqrt{1+9+25}=\sqrt{35} \\ c=AB=\sqrt{1+4+36}=\sqrt{41} \end{gathered}$$

Observing the above sides, ${\left(AB\right)}^2={\left(BC\right)}^2+{\left(CA\right)}^2$

The above traingle is rightangled triangle.

The circumecentre of the triangle is midpoint of the hypotenuse, The circum radius is half of hypotenuse

Hence, the circum radius is $\frac{\sqrt{41}}{2}$

\