Answer :

Let points be


P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)


Calculating PQ


P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)


Distance PQ


Here,


x1 = 0, y1 = 7, z1 = 10


x2 = – 1, y2 = 6, z2 = 6


Distance PQ





Calculating QR


Q ≡ (– 1, 6, 6) and R ≡ (– 4, 9, 6)


Distance QR


Here,


x1 = – 1, y1 = 6, z1 = 6


x2 = – 4, y2 = 9, z2 = 6


Distance QR





Calculating PR


P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)


Distance PR


Here,


x1 = 0, y1 = 7, z1 = 10


x2 = – 4, y2 = 9, z2 = 6


Distance PR





Now,


PQ2 + QR2 = 18 + 18 = 36 = PR2


Hence, According to converse of pythagoras theorem,


the given vertices P, Q & R are the vertices of a right – angled triangle at Q.


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