Answer
The particles do not collide but their paths intersect at $(1,1,1) $ & $(2,4,8) $
Work Step by Step
Since, $t=1+2t; t^2=1+6t; t^3=1+14t$
This gives that the first equation is satisfied for $t=-1$
Further, $t_1=1+2t_2; t_1^2=1+6t_2; t_1^3=1+14t_2$
Further, $(1+2t_2)^2=1+6t_2 \implies 4t_2+4t_2^2=6t_2$
This gives: $t_2=0 $ or, $t_2=\dfrac{1}{2}$
Set $t_2=0 $ then we have $t_2=\dfrac{1}{2}$,
Then, we have $t_1=1+2(0)=1$
and $t_2=\dfrac{1}{2}$, then we get $t_1=1+\dfrac{1}{2} \times 2=2$
Two point of intersections are as follows:
a) when $t_1=1 \implies (x,y,z) =(t_1,t_2^2,t_1^3)=(1,1,1)$
b) when $t_1=2\implies (x,y,z) =(t_1,t_2^2,t_1^3)=(2,4,8)$
Hence, we get the particles do not collide but their paths intersect at $(1,1,1) $ & $(2,4,8) $
