Answer
The particles do not collide but their paths intersect at $(1,1,1) , (2,4,8) $.
Work Step by Step
Consider $t=1+2t; t^2=1+6t; t^3=1+14t$
Let the first equation can be satisfied for $t=-1$
Further, $t_1=1+2t_2 \\t_1^2=1+6t_2 \\ t_1^3=1+14t_2$
Now, we have $(1+2t_2)^2=1+6t_2 \\4t_2+4t_2^2=6t_2$
This gives: $t_2=0 $ or, $t_2=\dfrac{1}{2}$
When $t_2=0 $; $t_1=1+2(0)=1$
and $t_2=\dfrac{1}{2}$; $t_1=1+2(0.5)=2$
Thus, we get the two points of intersection:
$(x,y,z) =(t_1,t_2^2,t_1^3)=(1,1,1)$ for $t_1=1$
$(x,y,z) =(t_1,t_2^2,t_1^3)=(2,4,8) $ for $t_1=2$
Thus, the particles do not collide but their paths intersect at $(1,1,1) , (2,4,8) $
