91Ó°ÊÓ

To find the distance between two points in 3-dimensional space, you can use the distance formula: $$distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$ Where \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the coordinates of the two points you want to find the distance between. Calculate the distances between points A, B, and C: AB distance: $$AB = \sqrt{(7-5)^2 + (16-6)^2 + (4-2)^2}$$ $$AB = \sqrt{2^2 + 10^2 + 2^2} = \sqrt{4+100+4} = \sqrt{108}$$ AC distance: $$AC = \sqrt{(6-5)^2 + (7-6)^2 + (3-2)^2}$$ $$AC = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1+1+1} = \sqrt{3}$$ BC distance $$BC = \sqrt{(6-7)^2 + (7-16)^2 + (3-4)^2}$$ $$BC = \sqrt{1^2 + (-9)^2 + (-1)^2} = \sqrt{1+81+1} = \sqrt{83}$$ Now that we have the lengths of all three sides of the triangle, we can move on to the next step.

To find the semi-perimeter of the triangle, add the lengths of all three sides and divide by 2: $$s = \frac{AB + AC + BC}{2}$$ Plug in the values we found in step 1: $$s = \frac{\sqrt{108} + \sqrt{3} + \sqrt{83}}{2}$$ Keep the expression in this form for now, as it will make calculations easier in the next step.

Heron's formula for calculating the area of a triangle is as follows: $$Area = \sqrt{s(s - a)(s - b)(s - c)}$$ Where \(a, b, c\) are the side lengths of the triangle, and \(s\) is the semi-perimeter calculated in step 2. Plug in the values: $$Area = \sqrt{(\frac{\sqrt{108} + \sqrt{3} + \sqrt{83}}{2})(\frac{\sqrt{108} + \sqrt{3} + \sqrt{83}}{2} - \sqrt{108})(\frac{\sqrt{108} + \sqrt{3} + \sqrt{83}}{2} - \sqrt{3})(\frac{\sqrt{108} + \sqrt{3} + \sqrt{83}}{2} - \sqrt{83})}$$ Calculating this expression with a calculator gives: $$\approx 3.74$$

The area of the triangle with vertices A(5, 6, 2), B(7, 16, 4), and C(6, 7, 3) is approximately 3.74 square units.