= Solution
Put $\delta=k_Ml$. Eliminating the middle-segment <complex amplitudes> using the right-hand <boundary conditions> gives
$$
1+r=t\left(\cos\delta-i\frac{Z_R}{Z_M}\sin\delta\right),\qquad
Z_L(1-r)=t\left(Z_R\cos\delta-iZ_M\sin\delta\right).
$$
Hence the displacement <reflection coefficient> is
$$
r=\frac{(Z_L-Z_R)\cos\delta+i(Z_M-Z_LZ_R/Z_M)\sin\delta}
{(Z_L+Z_R)\cos\delta-i(Z_M+Z_LZ_R/Z_M)\sin\delta}.
$$
For finite positive real <mechanical impedances>, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:
$$
(Z_L-Z_R)\cos\delta=0,\qquad (Z_M^2-Z_LZ_R)\sin\delta=0.
$$
This gives all three possibilities.
If the exterior strings differ, $Z_L\ne Z_R$, then $\cos\delta=0$, so $\sin\delta\ne0$ and
$$
\boxed{Z_M=\sqrt{Z_LZ_R},\qquad l=\frac{(2n+1)\lambda_M}{4},\quad n=0,1,2,\ldots.}
$$
The middle segment is a <quarter-wave impedance transformer>. Its two reflected contributions cancel at these odd quarter-<wavelength> thicknesses. Here $\lambda_M=2\pi/k_M$ is specifically the <wavelength> in the middle string; the matching is not generally valid at every <frequency>.
If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin\delta=0$, giving
$$
\boxed{l=\frac{n\lambda_M}{2},\quad n=1,2,\ldots.}
$$
A whole number of half-<wavelengths> reproduces the exterior <mechanical impedance> at the input. The formal $n=0$ limit is zero <frequency> or zero segment length, rather than a finite-<wavelength> resonance.
Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted <power>.
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