the hypotheses, it is clear that if $\widetilde{\theta}=e^{\frac{\left(
f+2u\right)  }{3}}\theta$ is a pseudo-Einstein contact form, then$$\gamma=0.$$ Hence we can solve the inhomogeneous tangential Cauchy-Riemann equation $$\overline{\partial}_{b}\varphi=\sigma_{\overline{1}}\theta^{\overline{1}}$$ by Lemma \[l3\]. Note that this implicitly implies $f$ is CR-pluriharmonic. So the sufficient part is completed.

As for the necessary part, it’s obvious from Theorem \[t1\].

Before to go further, we need the following key lemma.

\[l5\] Let $(M,J,\theta)$ be a closed strictly pseudoconvex CR $3$-manifold with $c_{1}(T_{1,0}M)=0.$ With the notations as above, the following equality holds$$\int_{M}\left(  R-\frac{1}{2}Tor-\frac{1}{2}Tor^{\prime}\right)  \left(
\gamma,\gamma \right)  d\mu+\int_{M}\left \vert \gamma_{1,1}\right \vert ^{2}d\mu+\int_{M}Qud\mu+\int_{M}\left(  P_{0}u^{\bot}\right)  u^{\bot}d\mu=0.\label{2018BB}$$

From the equality $\left(  \ref{0}\right)  $, we are able to get$$\begin{array}
[c]{ccl}\left(  R_{,1}-iA_{11,\overline{1}}\right)  \gamma_{\overline{1}} & = &
W_{1}\gamma_{\overline{1}}\\
& = & 2\left(  u_{\overline{1}11}+iA_{11}u_{\overline{1}}\right)
\gamma_{\overline{1}}+iA_{11}\gamma_{\overline{1}}\gamma_{\overline{1}}-i\gamma_{1,0}\gamma_{\overline{1}}\\
& = & 2\left(  u_{\overline{1}11}+iA_{11}u_{\overline{1}}\right)
\gamma_{\overline{1}}+iA_{11}\gamma_{\overline{1}}\gamma_{\overline{1}}-\left(  \gamma_{1,1\overline{1}}-R\gamma_{1}\right)  \gamma_{\overline{1}}.
\end{array}$$

Taking the integration over $M$ of both sides and its conjugation, we have, by the fact that $\gamma_{1,\overline{1}}=0$, $$i\int_{M}\left(  A_{11,\overline{1}}\gamma_{\overline{1}}-A_{\overline
{1}\overline{1},1}\gamma_{1}\right)  d\mu+\int_{M}\left(  2R-Tor\right)
\left(  \gamma,\gamma \right)  d\mu+2\int_{M}\left \vert \gamma_{1,1}\right \vert
^{2}d\mu-2\int_{M}Tor\left(  d_{b}u,\gamma \right)  d\mu=0.$$ That is $$\int_{M}\left(  R-\frac{1}{2}Tor-\frac{1}{2}Tor^{\prime}\right)  \left(
\gamma,\gamma \right)  d\mu+\int_{M}\left \vert \gamma_{1,1}\right \vert ^{2}d\mu-\int_{M}Tor\left(  d_{b}u,\gamma \right)  d\mu=0.\label{34}$$ On the other hand, it follows from the equality $\left(  \ref{0}\right)  $ that $$\left(  R,_{1}-iA_{11},_{\overline{1}}\right)  u_{\overline{1}}=W_{1}u_{\overline{1}}=\left[  2P_{1}u+i\left(  A_{11}\gamma_{\overline{1}}-\gamma_{1,0}\right)  \right]  u_{\overline{1}}.\label{30A}$$ By the fact 