= Solution
For two independent left-handed <Weyl spinors> $\chi$ and $\eta$, a <Dirac mass term> is
$$
\boxed{\mathcal L_D=-m\chi^\alpha\eta_\alpha-m^*\bar\chi_{\dot\alpha}\bar\eta^{\dot\alpha}.}
$$
The four-component <Dirac field> $\Psi_D=(\chi_\alpha,\bar\eta^{\dot\alpha})^T$ has independent particle and antiparticle degrees of freedom; after a phase redefinition making $m$ real, the same term is $-m\bar\Psi_D\Psi_D$. The two left-handed spinors can carry opposite charges under a conserved <U(1)> symmetry, $\chi\mapsto e^{iq\omega}\chi$, $\eta\mapsto e^{-iq\omega}\eta$. A gauge-invariant <Dirac mass term> contracts conjugate <gauge group representations>.
For one left-handed <Weyl spinor>, a <Majorana mass term> instead has the form
$$
\boxed{\mathcal L_M=-\frac12\left(m\chi^\alpha\chi_\alpha+m^*\bar\chi_{\dot\alpha}\bar\chi^{\dot\alpha}\right).}
$$
Its self-contraction need not vanish, by the <Weyl spinor bilinear exchange identity>. The factor $1/2$ avoids double counting identical fields in variation. The associated four-component <Majorana fermion> $\Psi_M=(\chi_\alpha,\bar\chi^{\dot\alpha})^T$ satisfies $\Psi_M^c=\Psi_M$, so its particle is its own antiparticle. A <Majorana mass term> breaks continuous fermion-number phase symmetry for $\chi$, and is allowed only if its gauge indices admit an invariant symmetric contraction, as for a neutral field or a real <gauge group representation>. An isolated field of nonzero <U(1)> charge cannot have it. More generally, the mass matrix of several left-handed <Weyl spinors> is symmetric, since $\chi_i\chi_j=\chi_j\chi_i$.
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