Có bao nhiêu số phức $z$ thỏa mãn $3\left| z+\overline{z}...

Đặt $z=x+iy$ với $x,y\in \mathbb{R}$.
$\left| z+2-3i \right|=\left| \overline{z}-4+i \right|$ $\Leftrightarrow {{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( x-4 \right)}^{2}}+{{\left( -y+1 \right)}^{2}}$ $\Leftrightarrow y=3x-1$
$3\left| z+\overline{z} \right|+2\left| z-\overline{z} \right|=12$ $\Leftrightarrow 3.\left| 2x \right|+2.\left| 2iy \right|=12$ $\Leftrightarrow 3\left| x \right|+2\left| y \right|=6$ $\Rightarrow 3\left| x \right|+2\left| 3x-1 \right|=6$
$\Leftrightarrow \left[ \begin{matrix}
\left\{ \begin{aligned}
& x<0 \\
& -3x+2\left( 1-3x \right)=6 \\
\end{aligned} \right. \\
\left\{ \begin{aligned}
& 0\le x<\dfrac{1}{3} \\
& 3x+2\left( 1-3x \right)=6 \\
\end{aligned} \right. \\
\left\{ \begin{aligned}
& x\ge \dfrac{1}{3} \\
& 3x+2\left( 3x-1 \right)=6 \\
\end{aligned} \right. \\
\end{matrix} \right.\Leftrightarrow \left[ \begin{matrix}
\left\{ \begin{aligned}
& x<0 \\
& x=-\dfrac{4}{9} \\
\end{aligned} \right. \\
\left\{ \begin{aligned}
& 0\le x<\dfrac{1}{3} \\
& x=\dfrac{-4}{3} \\
\end{aligned} \right. \\
\left\{ \begin{aligned}
& x\ge \dfrac{1}{3} \\
& x=\dfrac{8}{9} \\
\end{aligned} \right. \\
\end{matrix} \right.\Leftrightarrow \left[ \begin{aligned}
& x=-\dfrac{4}{3} \\
& x=\dfrac{8}{9} \\
\end{aligned} \right.$
Vậy có 2 số phức $z$ thỏa yêu cầu bài toán.