The Petersen graph has ten vertices, is cubic, has girth five and has $\kappa =\lambda =3$

The vertex set has $\left(\genfrac{}{}{0px}{}{5}{2}\right)=10$ elements by [L1]. For a vertex $A$, its neighbours are the two-element subsets of the three-element complement $\mathrm{\Omega }\setminus A$, so it has $\left(\genfrac{}{}{0px}{}{3}{2}\right)=3$ neighbours. Thus $G$ is cubic and $\delta (G)=3$.

If distinct vertices $A,B$ are disjoint, their union has four elements and there is no two-element subset disjoint from both. If they intersect, their union has three elements and its two-element complement is their unique common neighbour.

Deleting any two vertices leaves a connected graph. Up to relabelling $\mathrm{\Omega }$, two deleted two-element subsets are either the disjoint pair $12,34$ or the intersecting pair $12,13$. In the first case the remaining vertices have the spanning path $13,25,14,35,24,15,23,45$; in the second they have the spanning path $24,35,14,25,34,15,23,45$. Every successive pair displayed is disjoint. Relabelling preserves disjointness, so these cases cover every pair.

There is no triangle: adjacent vertices are disjoint and have no common neighbour by step 1.2. There is no $4$-cycle, because its two opposite vertices would have two distinct common neighbours, also impossible by step 1.2.

Every two distinct vertices are at distance at most $2$: disjoint ones are adjacent, and intersecting ones have the common neighbour from step 1.2. Hence $G$ is connected.

The vertices $12,34,15,23,45$ in that order form a $5$-cycle, since consecutive sets, including $45$ and $12$, are disjoint. With step 2.1 and [F2], this proves $g(G)=5$.

No set of at most two edges disconnects $G$. Indeed, if such a deletion produced a component on a smallest vertex set $S$, then $1\le \mathrm{\mid }S\mathrm{\mid }\le 5$. Writing $e(S)$ for the number of edges inside $S$, cubicity gives $3\mathrm{\mid }S\mathrm{\mid }-2e(S)$ edges leaving $S$. For $\mathrm{\mid }S\mathrm{\mid }=1,2,3$ this number is at least $3,4,5$, since a simple triangle-free graph on those sizes has at most $0,1,2$ edges. For $\mathrm{\mid }S\mathrm{\mid }=4$, the absence of triangles and $4$-cycles makes the induced graph acyclic, so it has at most $3$ edges and at least $6$ edges leave. For $\mathrm{\mid }S\mathrm{\mid }=5$, it is either acyclic with at most $4$ edges or is a chordless $5$-cycle with $5$ edges, so at least $5$ edges leave. Every case has at least three boundary edges, a contradiction. Therefore $\lambda (G)\ge 3$.

Deleting one vertex also leaves a connected graph: choose a second vertex, use step 1.3 on the remaining eight, and then add the second vertex back; it retains at least two of its three neighbours. Step 2.2 covers deletion of no vertices. Thus no vertex cut has size at most $2$, and $\kappa (G)\ge 3$.

By [L2] and step 1.1, $\lambda (G)\le \delta (G)=3$. Together with step 3.2 this gives $\lambda (G)=3$.

Whitney's inequality [L2] and step 4.1 give $\kappa (G)\le 3$, while step 3.3 gives the reverse bound. Hence $\kappa (G)=\lambda (G)=3$; both lower bounds were obtained in steps 3.2 and 3.3, which use the cubic regularity of $G$.