Vector factorisation - a variation

If we expand all the dot products and add them, we get $ad+ae+af+bd+be+bf+cd+ce+cf=221$ . The left-hand side of this factorises!

We have $(a+b+c)(d+e+f)=221=13\cdot 17$ . Now, since $a,b,c,d,e,f$ are all positive integers, both $a+b+c$ and $d+e+f$ are greater than $1$ ; so the only possibility is that $a+b+c=13$ and $d+e+f=17$ (or vice-versa), giving $a+b+c+d+e+f=\boxed{30}$ .

Note we're safe to do this since the problem tells us such a solution exists. It's a harder problem to find the solution, but that isn't needed to solve the problem. However, one such solution is $(a,b,c,d,e,f)=(2,4,7,5,9,3)$ .