Two dense lobes meeting at one cut vertex give $\kappa(G)=1<\lambda(G)=2<\delta(G)=3$

The graph is connected, and deleting $v$ separates the two lobes. No deletion of zero vertices disconnects a connected graph, so $\kappa(G)=1$.

In each lobe, $a_i$ and $b_i$ have two neighbours inside the lobe and the neighbour $v$, so degree $3$; $c_i,d_i$ have degree $3$ inside the lobe; and $v$ has degree $4$. Hence $\delta(G)=3$.

Deleting $va_1$ and $vb_1$ separates the first lobe from the rest, so $\lambda(G)\le2$.

Every edge lies on a cycle. The edges $va_i$ and $vb_i$ lie on the $4$-cycle $v,a_i,c_i,b_i,v$. Each internal edge incident with $a_i$ lies on the triangle $a_i,c_i,d_i,a_i$, and each internal edge incident with $b_i$ lies on $b_i,c_i,d_i,b_i$. Thus deleting one edge leaves an alternate path between its endpoints and cannot disconnect the graph, so $\lambda(G)\ge2$.

Steps 1.3 and 1.4 give $\lambda(G)=2$. Together with steps 1.1 and 1.2, this proves $\kappa(G)=1<\lambda(G)=2<\delta(G)=3$ and refutes equality in both Whitney inequalities.