= Solution
Use signature $(-,+,+,+)$, $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$, and unit-weight antisymmetrized gamma products. The massless <Rarita-Schwinger field> has the gauge freedom
$$
\psi_\mu\longmapsto\psi_\mu+\partial_\mu\epsilon,
$$
since the antisymmetric $\gamma^{\mu\nu\rho}$ annihilates $\partial_\nu\partial_\rho\epsilon$. This gauge freedom is essential to remove the lower-spin components of the vector-spinor.
Put $\chi=\gamma^\mu\psi_\mu$. Contracting the field equation with $\gamma_\mu$ and using $\gamma_\mu\gamma^{\mu\nu\rho}=2\gamma^{\nu\rho}$ gives
$$
\not\partial\chi-\partial^\mu\psi_\mu=0.
$$
Expanding the original three-gamma product then reduces its equation to
$$
\not\partial\psi^\mu-\partial^\mu\chi=0.
$$
A gauge transformation changes $\chi$ by $\not\partial\epsilon$. Locally solve $\not\partial\epsilon=-\chi$ to impose gamma-trace gauge. The two equations become
$$
\boxed{\gamma\cdot\psi=0,\qquad\partial\cdot\psi=0,\qquad\not\partial\psi_\mu=0.}
$$
Applying $\not\partial$ again gives $\Box\psi_\mu=0$. Nonzero propagating Fourier modes therefore have $p^2=0$: \b[the physical modes are massless].
For the <massless Rarita-Schwinger polarization count>, choose a nonzero future null momentum $p^\mu=(\omega,0,0,\omega)$. The remaining gauge parameters satisfy $\not p\epsilon=0$. Transversality gives $\psi_0+\psi_3=0$, and this residual gauge removes $\psi_0$, hence also $\psi_3$, because $\psi_0$ itself satisfies $\not p\psi_0=0$. Only $\psi_1,\psi_2$ remain, with
$$
\not p\psi_1=\not p\psi_2=0,\qquad
\gamma^1\psi_1+\gamma^2\psi_2=0,\qquad
\psi_2=\gamma^1\gamma^2\psi_1.
$$
The kernel of $\not p$ on four-component spinors has complex dimension two, so there are exactly two independent polarization amplitudes.
To identify their <helicities>, the spinor rotation generator about the third axis in this convention is $S_3=-i\gamma^1\gamma^2/2$. Its two eigenvalues on that kernel are $h_s=\pm1/2$. If $\psi_1$ is such an eigenvector, then $\psi_2=2ih_s\psi_1$. Its transverse vector polarization is consequently $(1,+i)$ for $h_s=+1/2$ or $(1,-i)$ for $h_s=-1/2$. These carry vector helicity $+1$ or $-1$, respectively. Adding the vector and spinor angular momenta leaves
$$
\boxed{h=+\frac32,\quad h=-\frac32,\qquad\text{two physical helicity states}.}
$$
The unwanted opposite-sign vector-spinor combinations, which would have helicity $\pm1/2$, violate the gamma-trace constraint. For the <gravitino> of simple <supergravity>, its <Majorana spinor> reality condition identifies the negative-frequency antiparticle modes, giving two propagating degrees of freedom. If one instead chooses a complex Dirac vector-spinor, its independent antiparticle sector doubles the particle-plus-antiparticle count; the equation alone does not impose a Majorana condition.
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