**Given:** A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

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

**To find:** the ratio in which C divides AB

**Formula used: **

**Section Formula:**

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

\(\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)\)

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates

Therefore, m = k and n = 1

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

Coordinates of C using section formula:

⇒ (-4,1,-10) = \(\Big(\frac{k(-1)+1(2)}{k+1},\frac{k(2)+1(3)}{k+1},\frac{k(-3)+1(4)}{k+1}\Big)\)

⇒ (-4,1,-10) = \(\Big(\frac{-k+2}{k+1},\frac{2k+3}{k+1}, \frac{-3k+4}{k+1}\Big)\)

On comparing:

\(\frac{-k+2}{k+1}\) = - 4

⇒ -k + 2 = -4(k + 1)

⇒ -k + 2 = -4k – 4

⇒ 4k – k = - 2 – 4

⇒ 3k = - 6

⇒ k = \(\frac{-6}{3}\)

⇒ k = - 2

\(\frac{2k+3}{k+1}\) = 1

⇒ 2k + 3 = k + 1

⇒ 2k – k = 1 – 3

⇒ k = – 2

\(\frac{-3k+4}{k+1}\) = -10

⇒ -3k + 4 = -10(k + 1)

⇒ -3k + 4 = -10k – 10

⇒ -3k + 10k = -10 – 4

⇒ 7k = -14

⇒ k = \(\frac{-14}{7}\)

⇒ k = - 2

The value of k is the same in all three times

Hence, A, B and C are collinear

As k = -2

C divides AB externally in ratio 2:1