Answer
They do not collide but their paths intersect at $(1,1,1)$ and $(2,4,8)$
Work Step by Step
$t=1+2t; t^2=1+6t; t^3=1+14t$
The first equation is satisfied for $t=-1$
Further, $t_1=1+2t_2; t_1^2=1+6t_2; t_1^3=1+14t_2$
Now, $(1+2t_2)^2=1+6t_2$
or, $4t_2+4t_2^2=6t_2$
or, $t_2=0 $ or, $t_2=\dfrac{1}{2}$
When $t_2=0 $ or, $t_2=\dfrac{1}{2}$, then $t_1=1+2(0)=1$
and $t_2=\dfrac{1}{2}$, then $t_1=1+2(1/2)=2$
Now, the two points of intersections are:
$(x,y,z) =(t_1,t_2^2,t_1^3)=(1,1,1)\ when\ t_1=1$
and
$(x,y,z) =(t_1,t_2^2,t_1^3)=(2,4,8)\ when\ t_1=2$
This means that the particles do not collide but their paths intersect at $(1,1,1)$ and $(2,4,8)$
