Prove that ABCD is a rectangle if A (-3; 2; 2). B (-1; -8; 13). C (-15; -13; 11). D (-17; -3; 0)

In order to prove that ABCD is a rectangle, it is enough to find the lengths of all sides:
AB = {2; -ten; 11}; | AB | = (4 + 100 + 121) ^ (1/2) = 15;
BC = {-14; -five; -2}; | BC | = (196 + 25 + 4) ^ (1/2) = 15;
AD = {-14; -five; -2}; AD | = (196 + 25 + 4) ^ (1/2) = 15;
CD = {-2; ten; -11}; | CD | = (4 + 100 + 121) ^ (1/2) = 15.
Let’s find the cosines:
cos (AB; BC) = (- 28 + 50-22) / 225 = 0;
cos (BC; CD) = (28-50 + 22) / 225 = 0;
cos (CD; AD) = (28-50 + 22) / 225 = 0;
cos (AD; AB) = (- 28 + 50-22) / 225 = 0.
We got that all sides are equal, all angles are 90, then we have a square.