 Solution

Use the ordinary treewidth of the primal graph of the constraint hypergraph: two variables are adjacent whenever they occur together in an all-different constraint. Equivalently, a tree decomposition of this constraint hypergraph has a bag containing every constraint scope, together with the running-intersection requirement for each variable. This is distinct from taking the treewidth of its incidence drawing.

Introduce a central bag $B_0=\{2,4,5\}$ and the five scope bagsJoin $B_0$ to each of $B_1,B_2,B_3$, join $B_3$ to $B_4$, and join $B_1$ to $B_5$:

Every constraint scope is contained in its own bag, and every variable occurs. The nontrivial running-intersection checks are: $2$ occurs along $B_1-B_0-B_2$; $4$ along the connected branches $B_5-B_1-B_0-B_3$; $5$ along $B_2-B_0-B_3$; $1$ along $B_1-B_5$; and $6$ along $B_3-B_4$. Each of the other variables occurs in just one bag. This is therefore a valid tree decomposition of width $3-1=2$.

Conversely, any of the original three-variable constraint scopes forms a clique of size three in the primal graph. Such a clique must be contained in one bag of every tree decomposition, forcing width at least two. The upper and lower bounds agree: