2-Beltrami derivative of the normal vector of surface in $\textbf{R}^3$

Question

Assume a surface $\textbf{S}\in C^3$, $\overline{x}=\overline{x}(u,v)$, $(u,v)\in D$ and a function $f(u,v)\in C^2$ defined on $D$. Moreover let $\{\overline{e}_1,\overline{e}_2,\overline{e}_3\}$ be an orthonormal moving frame of $\textbf{S}$ and $\omega_1$, $\omega_2$ are the differential forms related with the surface $\textbf{S}$ (we assuming that $\overline{e}_3=\overline{n}$ is the normal vector to the tangent plane of $\textbf{S}$): $$ d\overline{x}=\omega_1\overline{e}_1+\omega_2\overline{e}_2.\tag 1 $$ For any such $f=f(u,v)$ with
$$ df=\partial_1fdu+\partial_2fdv=\nabla_1f\omega_1+\nabla_2f\omega_2, $$ $\nabla_1f$ and $\nabla_2f$ are the Pfaff derivatives of $f$ defined from $\textbf{S}$. Also it is known that if $$ q_1=\frac{d\omega_1}{\omega_1\wedge\omega_2}\textrm{, }q_2=\frac{d\omega_2}{\omega_1\wedge\omega_2}, $$ then $$ \nabla_1\nabla_2f-\nabla_2\nabla_1f+q_1\nabla_1f+q_2\nabla_2f=0\textrm{, (condition)} $$ and the 2-Beltrami operator, for a function $f$ or a vector field $\overline{F}$ of class $C^2$ is $$ \Delta_2f:=\nabla_1\nabla_1f+\nabla_2\nabla_2f+q_2\nabla_1f-q_1\nabla_2f,\tag 2 $$ and $$ \Delta_2\overline{F}:=\nabla_1\nabla_1\overline{F}+\nabla_2\nabla_2\overline{F}+q_2\nabla_1\overline{F}-q_1\nabla_2\overline{F}. $$ Prove that: $$ \frac{1}{2}\Delta_2\overline{n}=\overline{\textrm{grad}H}+(K-2H^2)\overline{n},\tag 3 $$ where $K$ and $H$ are the Gauss curvature and the mean curvature resp of the surface $\textbf{S}$. The operator $\overline{\textrm{grad}f}$ is the gradient of $f$ and it is defined for any function $f\in C^2$ as $$ \overline{\textrm{grad}f}:=(\nabla_1f)\overline{e}_1+(\nabla_2f)\overline{e}_2. $$