The line from (a,b,c) to (x,y,z) in R^3 is given by (a,b,c) + t*(x-a, y-b, z-c), with t in [0,1]. Since you need the parameter in [0,1] for the three poligonal lines you need to "scale" it:

== (a,b,c)) + 3t*((x-a), (y-b), (z-c)), t in [0,1/3] is from (a,b,c) to (x,y,z) ==> in your case we get:

(0,0,0) + 3t(2, 1, 1) = {(6t, 3t, 3t) , t in [0,1/3]}

== (x,y,z) + (3t-1)(p-x, q-y,r-z), with t in [1/3,2/3] is from (x,y,z) to (p,q,r), so in your case is...etc

Tonio

==