Given: Points are A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) 

To prove: the triangle formed by given points is an isosceles right-angled triangle Isosceles right-angled triangle is a triangle whose two sides are equal and also satisfies Pythagoras Theorem 

Formula used: 

The distance between any two points (a, b, c) and (m, n, o) is given by,

 \(\sqrt{(a−m)^2+(b−n)^2+(c−0)^2}\)

Therefore, 

Distance between A(0, 7, 10) and B(-1, 6, 6) is AB,

= \(\sqrt{0-(-1))^2+(7-6)^2+(10-6)^2} \) 

= \(\sqrt{1^2+1^2+4^2} \) 

 = \(\sqrt{1+1+16} \)

= \(\sqrt{18}\)

 = \(3\sqrt{2}\)

Distance between B(-1, 6, 6) and C(-4, 9, 6) is BC,

 = \(\sqrt{(-1-(-4))^2+(6-9)^2+(6-6)^2} \) 

= \(\sqrt{3^2+(-3)^2+0^2} \) 

 = \(\sqrt{9+9} \)

= \(\sqrt{18}\) 

= \(3\sqrt{2}\)

Distance between A(0, 7, 10) and C(-4, 9, 6) is AC,

 = \(\sqrt{(0-(-4))^2+(7-9)^2+(10-6)^2} \) 

= \(\sqrt{(4)^2+(-2)^2+4^2} \) 

 = \(\sqrt{16+4+16} \)

= \(\sqrt{36}\)  

= 6

Since, AB = BC 

AB² + BC²

= \((3\sqrt{2})^2+(3\sqrt{2})^2\)  

= 18 + 18

= 36 = AC² 

As, AB = BC and AB² + BC² = AC² 

Thus, Δ ABC is an isosceles-right angled triangle