= Solution
We construct a family by <seven-triangle Dirichlet transplantation>. Start with a scalene <triangle> $\tau$ sufficiently close to an <equilateral triangle>. Its three side lengths are the three independent parameters; restrict to an open set where its angles $\alpha,\beta,\gamma$ are distinct and lie between $\pi/4$ and $\pi/2$. Label the sides $a,b,c$ opposite the corresponding vertices $A,B,C$.
For each of two <planar domains>, begin with tile $7$ and attach the remaining congruent copies by <Euclidean reflection> across the indicated sides. The attachment tables are
$$
\begin{array}{c|cc}
\text{side}&\text{point domain }\Omega_P&\text{line domain }\Omega_L\\\hline
a&(2,3),(6,7)&(1,3),(5,7)\\
b&(4,6),(5,7)&(2,6),(3,7)\\
c&(1,5),(3,7)&(4,5),(6,7)
\end{array}
$$
For example, $(6,7)$ in the first $a$ row means to reflect tile $7$ across its $a$ side to obtain tile $6$. Both attachment <graphs> are trees: there is a central tile with three arms, each consisting of two tiles. Every side not used in the table is an exterior boundary side.
At the <equilateral triangle>, these seven reflected tile interiors are disjoint; at each central vertex four sectors of angle $\pi/3$ meet, leaving an exterior gap $2\pi/3$. Other boundary vertices have angles $\pi/3$ or $2\pi/3$, and distinct nonincident pieces are separated. The positive angular gaps and separations persist under sufficiently small perturbations of the three side lengths. Thus this construction really gives connected polygonal <planar domains> on an open three-dimensional parameter set. Their only reentrant vertices are the central vertices $A,B,C$, with interior angles $4\alpha,4\beta,4\gamma$; all other boundary angles are one or two triangle angles and are less than $\pi$.
Because these three reentrant angles are distinct, an <isometry> of the two <planar domains> would send each labelled central vertex to its counterpart. A planar <isometry> is the restriction of a Euclidean rigid motion: its differential preserves the Euclidean connection, hence is constant on the connected domain. After aligning the central triangles, such a motion fixes three noncollinear points and is the identity. But the domains are not identical. On the arm attached to side $a$, the point construction attaches the last tile across side $b$, whereas the line construction attaches it across side $c$. The former side is therefore interior in one domain and exterior in the other. This proves \b[the two domains are nonisometric throughout this three-parameter family].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-22-propeller-drums.png]
{title=Two noncongruent seven-triangle domains with identical Dirichlet spectra; matching colours label matching sides}
We now prove equality of the <Dirichlet Laplacian> <spectra>, including multiplicities. Index $1,\ldots,7$ by the nonzero binary vectors $x=(x_1,x_2,x_3)$, with integer label $x_1+2x_2+4x_3$. The transformations
$$
s_a(x)=(x_1+x_2,x_2,x_3),\qquad
s_b(x)=(x_1,x_2+x_3,x_3),\qquad
s_c(x)=(x_1,x_2,x_3+x_1)
$$
are involutions over $\mathbb F_2$. Their point permutations $P_s$ are precisely the transpositions in the left attachment table, with the remaining points fixed. On covectors the inverse-transpose transformations give permutations $Q_s$ with exactly the right table. Let $B$ be the point-line <incidence matrix of a set system> of the <Fano plane>:
$$
B_{y,x}=\begin{cases}1,&y\cdot x=0,\\0,&y\cdot x=1.\end{cases}
$$
The incidence relation is invariant under transforming $x$ and its covector $y$ contragrediently, so $BP_s=Q_sB$ for all three sides. Each point lies on three lines, and two distinct points lie on exactly one common line. Consequently
$$
B^TB=2I+J,
$$
where $J$ is the all-ones <matrix>. Its <eigenvalues> are $9$ once and $2$ six times, so $B$ is invertible.
The <Dirichlet boundary condition> requires a sign change at an unglued side. Give each tile the sign $(-1)^{\text{distance from tile }7}$ and put
$$
D=\operatorname{diag}(1,1,-1,1,-1,-1,1).
$$
The signs alternate across each glued side in both domains. The signed side matrices are therefore
$$
S_s=-DP_sD,\qquad T_s=-DQ_sD.
$$
They have entry $+1$ across a glued pair and entry $-1$ on a fixed exterior tile. The invertible <seven-triangle Dirichlet transplantation> matrix $C=DBD$ satisfies
$$
\boxed{CS_s=T_sC\qquad(s=a,b,c).}
$$
This is an actual <Dirichlet boundary condition> intertwiner, not merely the unsigned incidence relation.
To justify the spectral conclusion even at the reentrant corners, use the closed <Dirichlet energy> form. Pull a function on either domain back to its seven reference triangles and write its components as a vector $F$. The form domain consists of $H^1$ components whose side traces agree on glued pairs and vanish on exterior sides; equivalently $F=S_sF$ on each side for the point domain, and $G=T_sG$ for the line domain. These trace conditions exactly characterize $H_0^1$ of the glued polygon. Both its $L^2$ norm and its <Dirichlet energy> are the sums of their seven reference-tile integrals.
Since the signed side matrices are symmetric, the intertwining identities also give $S_sC^TC=C^TCS_s$. Thus $(C^TC)^{-1/2}$ commutes with every $S_s$, and the polar factor
$$
U=C(C^TC)^{-1/2}
$$
is an orthogonal <matrix> with $US_s=T_sU$. Multiplication of the component vector by this constant <matrix> preserves the $L^2$ <inner product> and the integrated squared <gradients>, and maps the two form domains bijectively. The <self-adjoint operators> associated with these forms are consequently <unitarily equivalent>, so their <eigenvalues> and multiplicities agree. The <eigenfunctions> are smooth in the interiors; no unwarranted smoothness at polygonal corners is required. We have proved
$$
\boxed{\operatorname{Spec}_D(\Omega_P)=\operatorname{Spec}_D(\Omega_L),
\qquad\Omega_P\not\cong\Omega_L.}
$$
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