Solution (source code)

= Solution

The highest weight $(0,1,0)$ gives the six-dimensional second <exterior power> of the defining representation. Adding each unordered pair of the four defining weights from part i gives
$$
\begin{gathered}
(0,1,0),\qquad
(1,-1,1),\qquad
(1,0,-1),\\
(-1,0,1),\qquad
(-1,1,-1),\qquad
(0,-1,0).
\end{gathered}
$$
There are no multiplicities, in agreement with the assumption in the question.