Show that the points `A(1,2,7),B(2,6,3) and C(3,10,-10` are collinear.

Explanation

The gieven points are `A(1,2,7),B(2,6,3) and C(3,10,-1)`.
`therefore vec(AB)=PV of B-PV of A =(2hat(i)+6hat(j)+3hat(k))-(1hat(i)+2hat(j)+7hat(k))=(2-1)hat(i)+(6-2)hat(j)+(3-7)hat(k)=hat(i)+4hat(j)-4hat(k)`
Magnitude of `vec(AB),|vec(AB)|=sqrt(1)^2+(4)^2+(-4)^2)=sqrt(1+16+16)=sqrt(33)`
`vec(BC)=PV of C-PV of B=(3hat(i)+10hat(j)+1hat(k))-(hat(i)+6hat(j)+3hat(k))`.
`=(3-2)hat(i)+(10-6)hat(j)+(-1-3)hat(k)=hat(i)+4hat(j)-4hat(k)`
Magnitude of `vec(BC),|vec(BC)|=sqrt(1)^2+(4)^2+(-4)^2)=sqrt(1+16+16)=sqrt(33)`.
`vec(AC)=PV of C-PV of A=(3hat(i)+10hat(j)-1hat(k))-(1hat(i)+2hat(j)+7hat(k))`.
`=(3-1)hat(i)+(10-2)hat(j)+(-1-7)hat(k)=2hat(i)+8hat(j)-8hat(k)`
Magnitude of `vec(AC),|vec(AC)|=sqrt((2)^2+(8)^2+(8)^2)=sqrt(4+64+64)=sqrt(132)= 2sqrt(33)`
`=sqrt(33)+sqrt(33)`
`therefore |AC|=|AB|+|BC|`.
Hence , the given points A,B and C ARE coliinear.