Answer
Hence, the perimeter of the triangle is: $|AB|+|BC|+|CA|=\sqrt{17}+\sqrt{33}+6 \approx 15.87$
Work Step by Step
The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$
Let A, B, C be the points on a triangle.
We need to find the perimeter of the triangle: $|AB|+|BC|+|CA|$
Thus, $|AB|=\sqrt{(1+1)^2+(-1-2)^2+(3-1)^2}=\sqrt{17}$
Now, $|BC|=\sqrt{(3-1)^2+(4+1)^2+(5-3)^2}=\sqrt{33}$
and $|CA|=\sqrt{(-1-3)^2+(2-4)^2+(1-5)^2}=\sqrt{36}=6$
Hence, the perimeter of the triangle is: $|AB|+|BC|+|CA|=\sqrt{17}+\sqrt{33}+6 \approx 15.87$
