Show that the points (-2,3,5),(1,2,3) and (7,0,−1) are collinear.

If three points are collinear, then they lie on a line.

Firstly let us calculate distance between the 3 points

i.e. PQ, QR and PR

Calculating PQ

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

So here,

x₁ = – 2, y₁ = 3, z₁ = 5

x₂ = 1, y₂ = 2, z₂ = 3

Distance PQ = 

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

= \( \sqrt{(3)^2 + (-1)^2 + (-2)^2}\)

= \( \sqrt{9 + 1 + 4}\)

= \( \sqrt{14}\)

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

So here,

x₁ = 1, y₁ = 2, z₁ = 3

x₂ = 7, y₂ = 0, z₂ = – 1

Distance QR = 

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

= \( \sqrt{(6)^2 + (-2)^2 + (-4)^2}\)

= \( \sqrt{36 + 4 + 16}\)

= \( \sqrt{56}\)

= \( 2\sqrt{14}\)

Calculating PR

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

So here,

x₁ = – 2, y₁ = 3, z₁ = 5

x₂ = 7, y₂ = 0, z₂ = – 1

Distance PR =

\(\sqrt{(7 – (-2))^2 + (0 – 3)^2 + (-1 – 5)^2}\)

= \( \sqrt{(9)^2 + (-3)^2 + (-6)^2}\)

= \(\sqrt{81 + 9 + 36}\)

= \( \sqrt{126}\)

= \( 3\sqrt{14}\)

Thus, PQ = \( \sqrt{14}\), QR = \( 2\sqrt{14}\) and PR = \( 3\sqrt{14}\)

So, PQ + QR =\( \sqrt{14}+ 2\sqrt{14}\)\(\)

= \( 3\sqrt{14}\)

= PR

The points P, Q and R are collinear.