**Given:** A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

**To prove:** Points A, B and C are collinear

**Formula used: **

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB 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(3, -5, 1) and B(-1, 0, 8) is AB,

= \(\sqrt{(3-(-1))^2+(-5-0)^2+(1-8)^2}\)

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

= \(\sqrt{16+25+49}\)

= \(\sqrt{90}\)

= \(3\sqrt{10}\)

Distance between B(-1, 0, 8) and C(7, -10, -6) is BC,

= \(\sqrt{(-1-7)^2+(0-(-10))^2+(8-(-6))^2}\)

= \(\sqrt{(-8)^2+(10)^2+(14)^2}\)

= \(\sqrt{64+100+196}\)

= \(\sqrt{360}\)

= \(6\sqrt{10}\)

Distance between A(3, -5, 1) and C(7, -10, -6) is AC,

= \(\sqrt{(3-7)^2+(-5-(-10))^2+(-1-(-6))^2}\)

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

= \(\sqrt{16+25+49}\)

= \(\sqrt{90}\)

= \(3\sqrt{10}\)

Clearly,

AB + AC

= \(3\sqrt{10}+3\sqrt{10}\)

= \(6\sqrt{10}\)

= BC

Hence, A, B and C are collinear