Prove that the ponts `A(1,2,3),B(3,4,7),C(-3 ,-2, -5)` are collinear and find the ratio in which B divides AC.

Explanation

Clearly, `AB=(3-1)hati+(4-2)hatj+(7-3)hatk`
`=2hati+2hatj+4hatk`
and `BC=(-3-3)hati+(-2-4)hatj+(-5-7)hatk`
`=6hati-6hatj-12hatk`
`=-3(2hati+2hatj+4hatk)=-3AB`
`becauseBC=-3AB`
`therefore` A,B and C are collinear,
Now, let C diivide AB in the ratio k:1, then
`OC=(kOB+1*OA)/(k+1)`
`implies-3hati-2hatj-5hatk=(k(3hati+4hatj+7hatk)+(2hati+2hatj+3hatk))/(k+1)`
`implies-3hati-2hatj-5hatk=((3k+1)/(k+1))hati+((4k+2)/(k+1))hatj+((7k+3)/(k+1))hatk`
`implies(3k+1)/(k+1)=-3,(4k+2)/(k+1)=-2 and ((7k+3)/(k+1)=-5`
From, all relations, we get `k=(-2)/(3)`
Hence, C divides AB externally in the ratio 2:3.