If the points `A(2,1,1,),B(0,-1,4)andC(k,3,-2)` are collinear, then k= . . . . . . . .

Options

A. 0
B. 1
C. 4
D. `-4`

Explanation

(i) Given `a(2,1,1),b(0,-1,4), c(k,3-2)`
Since theses points are collinear
`rArr AB= lambda AC`
`b-a=lambda(c-a)`
`-2hati-2hatj+3hatk= lambda[(k-2)hati+2hatj-3hatk]`
On compaing both sides, we get
`2 lambda=-2`
`rArr lambda=-1`
`lambda(k-2)=-2`
`rArr -1(k-2)=-2`
`k-2=2`
`rArr k=4`
Hene, corret answer from the given alternatives is