Given: Points are A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) 

To prove: the quadrilateral formed by these 4 points is a square 

All sides of a square are equal 

Formula used: 

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

 \(\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}\)

Therefore, 

The distance between A(3, 3, 3) and B(0, 6, 3) is AB,

= \(\sqrt{(3-0)^2+(3-6)^2+(3-3)^2} \) 

= \(\sqrt{3^2+3^2+0^2} \) 

 = \(\sqrt{9+9} \)

= \(\sqrt{18}\)

 = \(3\sqrt{2}\)

Distance between B(0, 6, 3) and C(1, 7, 7) is BC,

 = \(\sqrt{(0-1)^2+(6-7)^2+(3-7)^2} \) 

= \(\sqrt{1^2+1^2+4^2} \) 

 = \(\sqrt{1+1+16} \)

= \(\sqrt{18}\) 

= \(3\sqrt{2}\)

Distance between C(1, 7, 7) and D(4, 4, 7) is CD,

 = \(\sqrt{(1-4)^2+(7-4)^2+(7-7)^2} \) 

= \(\sqrt{3^2+3^2+0^2} \) 

 = \(\sqrt{9+9+0} \)

= \(\sqrt{18}\)  

=\(3\sqrt{2}\)

The distance between A(3, 3, 3) and D(4, 4, 7) is AD,

= \(\sqrt{(3-4)^2+(3-4)^2+(3-7)^2}\)

= \(\sqrt{1+1+16}\)

= \(\sqrt{18}\) 

= \(3\sqrt{2}\)

Clearly, 

AB = BC = CD = AD 

Thus, Quadrilateral formed by ABCD is a square 

Hence Proved