**Given:** Points are A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4)

**To prove:** the quadrilateral formed by these 4 points is a rhombus All sides of both square and rhombus are equal But diagonals of a rhombus are not equal whereas they are equal for square

**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,

Distance between A(5, -1, 1) and B(7, -4, 7) is AB

= \(\sqrt{(5-7)^2+(-1-(-4))^2+(1-7)^2}\)

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

= \(\sqrt{4+9+36}\)

= \(\sqrt{49}\)

= 7

Distance between B(7, -4, 7) and C(1, -6, 10) is BC

= \(\sqrt{(7-1)^2+(-4-(-6))^2+(7-10)^2}\)

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

= \(\sqrt{36+4+9}\)

= \(\sqrt{49}\)

= 7

Distance between C(1, -6, 10) and D(-1, -3, 4) is CD,

= \(\sqrt{(1-(-1))^2+(-6-(-3))^2+(10-4)^2}\)

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

= \(\sqrt{4+9+36}\)

= \(\sqrt{49}\)

= 7

Distance between A(5, -1, 1) and D(-1, -3, 4) is AD,

= \(\sqrt{(5-(-2))^2+(1-(3))^2+(1-4)^2}\)

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

= \(\sqrt{36+4+9}\)

= \(\sqrt{49}\)

= 7

Clearly,

AB = BC = CD = AD

All sides are equal

Now, we will find length of diagonals

Distance between A(5, -1, 1) and C(1, -6, 10) is AC,

= \(\sqrt{(5-1)^2+(-1-(-6))^2+(1-10)^2}\)

= \(\sqrt{6^2+5^2+(-9)^2}\)

= \(\sqrt{36+25+81}\)

= \(\sqrt{142}\)

Distance between B(7, -4, 7) and D(-1, -3, 4) is BD,

= \(\sqrt{(7-(-1))^2+(-4-(-3))^2+(7-4)^2}\)

= \(\sqrt{8^2+(-1)^2+3^2}\)

= \(\sqrt{64+1+9}\)

= \(\sqrt{74}\)

Clearly,

AC ≠ BD

The diagonals are not equal, but all sides are equal

Thus, Quadrilateral formed by ABCD is a rhombus

Hence Proved