Show that the points A

Solution:

To prove: Points A, B, C form isosceles triangle.

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,

$\left(x_{1}, y_{1}, z_{1}\right)=(0,1,2)$

$\left(x_{2}, y_{2}, z_{2}\right)=(2,-1,3)$

$\left(x_{3}, y_{3}, z_{3}\right)=(1,-3,1)$

Length $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

$=\sqrt{(2-0)^{2}+(-1-1)^{2}+(3-2)^{2}}$

$=\sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}$

$=\sqrt{4+4+1}$

$=\sqrt{9}$

Length $B C=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

$=\sqrt{(1-2)^{2}+(-3+1)^{2}+(1-3)^{2}}$

$=\sqrt{(-1)^{2}+(-2)^{2}+(-2)^{2}}$

$=\sqrt{1+4+4}$

$=\sqrt{9}$

Length $A C=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}}$

$=\sqrt{(1-0)^{2}+(-3-1)^{2}+(1-2)^{2}}$

$=\sqrt{(1)^{2}+(-4)^{2}+(-1)^{2}}$

$=\sqrt{1+16+1}$

$=\sqrt{18}$

Also, $A B^{2}+B C^{2}=9+9=18=A C^{2}$

Therefore, points A, B, C forms an isosceles right-angled triangle