The vertices of the triangle are points A (10; -2; 8), B (8; 0; 7) C (10; 2; 8). calculate the perimeter of the triangle.

1. Find the sides of the triangle:

A (10; -2; 8);
B (8; 0; 7);
C (10; 2; 8);
AB = √ ((x (B) – x (A)) ^ 2 + ((y (B) – y (A)) ^ 2) + ((z (B) – z (A)) ^ 2);
AB = √ ((8 – 10) ^ 2 + (0 + 2) ^ 2 + (7 – 8) ^ 2) = √ ((- 2) ^ 2 + 2 ^ 2 + (-1) ^ 2) = √ (4 + 4 + 1) = √9 = 3;
AC = √ ((x (C) – x (A)) ^ 2 + ((y (C) – y (A)) ^ 2) + ((z (C) – z (A)) ^ 2);
AC = √ ((10 – 10) ^ 2 + (2 + 2) ^ 2 + (8 – 8) ^ 2) = √ (0 ^ 2 + 4 ^ 2 + 0 ^ 2) = √16 = 4;
BC = √ ((x (C) – x (B)) ^ 2 + ((y (C) – y (B)) ^ 2) + ((z (C) – z (B)) ^ 2);
BC = √ ((10 – 8) ^ 2 + (2 – 0) ^ 2 + (8 – 7) ^ 2) = √ (2 ^ 2 + 2 ^ 2 + 1 ^ 2) = √ (4 + 4 + 1) = √9 = 3.
2. The perimeter of the triangle is:

P (ABC) = AB + AC + BC;
P (ABC) = 3 + 4 + 3 = 10.
Answer: 10.