Show that D(-1, 4, -3) is the circumcentre of triangle ABC with vertices A(3, 2, -5), B(-3. 8, -5) and C(-3, 2, 1).
To prove: D is circumcenter of triangle ABC
Let us consider D as circumcenter of triangle ABC
∴ AD = BC = CD.
Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by
$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$
Here,
$\left(x_{1}, y_{1}, z_{1}\right)=(3,2,-5)$
$\left(x_{2}, y_{2}, z_{2}\right)=(-3.8,-5)$
$\left(x_{3}, y_{3}, z_{3}\right)=(-3,2,1)$
$\left(x_{4}, y_{4}, z_{4}\right)=(-1,4,-3)$
Length $A D=\sqrt{\left(x_{4}-x_{1}\right)^{2}+\left(y_{4}-y_{1}\right)^{2}+\left(z_{4}-z_{1}\right)^{2}}$
$=\sqrt{(-1-3)^{2}+(4-2)^{2}+(-3+5)^{2}}$
$=\sqrt{(-4)^{2}+(2)^{2}+(2)^{2}}$
$=\sqrt{16+4+4}$
$=\sqrt{24}$
Length $B D=\sqrt{\left(x_{4}-x_{2}\right)^{2}+\left(y_{4}-y_{2}\right)^{2}+\left(z_{4}-z_{2}\right)^{2}}$
$=\sqrt{(-1+3)^{2}+(4-8)^{2}+(-3+5)^{2}}$
$=\sqrt{(2)^{2}+(-4)^{2}+(2)^{2}}$
$=\sqrt{4+16+4}$
$=\sqrt{24}$
Length $C D=\sqrt{\left(x_{4}-x_{3}\right)^{2}+\left(y_{4}-y_{3}\right)^{2}+\left(z_{4}-z_{3}\right)^{2}}$
$=\sqrt{(-1+3)^{2}+(4-2)^{2}+(-3-1)^{2}}$
$=\sqrt{(2)^{2}+(2)^{2}+(-4)^{2}}$
$=\sqrt{4+4+16}$
$=\sqrt{24}$
Hence, the condition is consistent.
Hence, $D$ is circumcenter of triangle $A B C$.