Let $f:\mathbb{C}\rightarrow\mathbb{C}^3$ defined by $f=(e^{f_1},-e^{f_1},e^{f_3})$ where $f_1,f_3:\mathbb{C}\rightarrow\mathbb{C}$ are holomorphic.
We identify $\mathbb{C}^3$ to $\mathbb{R}^6$: ($z_1,z_2,z_3)=(x_1,y_1,x_2,y_2,x_3,y_3$). Let $H_1$, $H_2$, $H_3$ and $H_4$ be four hyperplanes in $\mathbb{C}^3$, defined by; $$\begin{array}{ccc} &H_1=&Span_\mathbb{R}\big[(1,0,0,0,0,0);(0,1,0,0,0,0)\big],\\ &H_2=&Span_\mathbb{R}\big[(0,0,1,0,0,0);(0,0,0,1,0,0)\big],\\ &H_3=& Span_\mathbb{R}\big[(0,0,0,0,1,0);(0,0,0,0,0,1)\big],\\ &H_4=&Span_\mathbb{R}\big[(1,0,1,0,1,0);(0,1,0,1,0,1)\big].\\ \end{array}$$ Question: Is there a real subspace $H$, of real dimension four such that $span_{\mathbb{R}}(H_i,H_ j, H^\perp)= \mathbb{R}^6$ for all $1\le i\neq j\le 4$ and such that $f$ avoids $H$?
I tried the four-dimensional subspace :
$$\textbf{(H)} \ \left\{\begin{array}{ccllll} X_1-X_2&=&0 & \\ X_1-X_3&=&0&\\ \end{array}\right. $$
$f$ avoid this subspace, but $H^\perp=Span_\mathbb{R}\big[(-1,0,-1,0,0,0);(-1,0,0,0,-1,0)\big]$ does not satifies the condition $Span_\mathbb{R}(H_i,H_j,H^\perp)=\mathbb{R}^6$ for all $j\neq k,~~j,k\in\lbrace 1,...,4\rbrace$
I tried also the following four-dimensional subspace: $$H=\left\{\begin{array}{cllll} 2y_1+x_2+y_2+y_3&=&0\\ ~~\\ x_1+2x_2+2x_3+y_2&=&0 \\ \end{array}\right.$$ Then $H^\perp=Span_\mathbb{R}\big[(0,2,1,1,0,1);(1,0,2,1,2,0)\big]$, which of course satisfies the condition $Span_\mathbb{R}(H^\perp,H_j^\perp,H_k^\perp)=\mathbb{R}^6$$Span_\mathbb{R}(H^\perp,H_j,H_k)=\mathbb{R}^6$ for all $j\neq k,~~j,k\in\lbrace 1,...,4\rbrace$, but $f$ does not avoid this subspace.
Please help and thanks in advance.
