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.