The procedure described in the books looks unnecessarily complicated.
The most natural thing (to me) is to use the formula
$$
\operatorname{Re}((a+ib)\overline{(u+iv)}) = au+bv = (a,b)\cdot (u,v)
$$
which expresses the dot product of real vectors in terms of complex numbers.
Here the conjugation can go over either factor:
$$
\operatorname{Re}(\overline{(a+ib)}{(u+iv)}) = \operatorname{Re}((a+ib)\overline{(u+iv)}) = (a,b)\cdot (u,v)
$$
In your problem, the expression $2x-y$ should be recognized as $(2,-1)\cdot (x,y)$. By the above, this is equal to $\operatorname{Re}(\overline{(2-i)}{(x+iy)})$. Hence the answer:
$$\operatorname{Re}((2+i)z) = 3$$
Which is what you have at the end, except you've made a typo replacing $2$ by $3$.
There is a really important aspect of complex numbers that depends on the complex plane having exactly this shape: complex multiplication.
Complex numbers can not only be characterized in cartesian coordinates by a real part and an imaginary part, but also in polar coordinates by a length and an angle.
You know that for any $z \in \mathbb{C}$ there exist $x, y \in \mathbb{R}$ such that $z = x+i\cdot y$, right? $x$ is the real part and $y$ is the imaginary part? Well, there also exist $r, \varphi \in \mathbb{R}$, $r \geq 0$ such that $z = r\cdot(\cos\varphi + i\sin\varphi)$. Here, $r$ is called the length or absolute value of $z$ and $\varphi$ is called the angle or argument, measured counterclockwise from the positive real axis.
We can use cartesian coordinates to add complex numbers: $$(x_1+i\cdot y_1) + (x_2+i\cdot y_2) = (x_1+x_2) + i\cdot(y_1+y_2)$$
We can use cartesian coordinates to multiply complex numbers:
$$(x_1+i\cdot y_1)\cdot (x_2+i\cdot y_2) = (x_1x_2-y_1y_2) + i\cdot(x_1y_2+y_1x_2)$$
However, we can also use polar coordinates to multiply complex numbers:
$$(r_1(\cos\varphi_1 + i\sin\varphi_1))\cdot(r_2(\cos\varphi_2 + i\sin\varphi_2)) = (r_1\cdot r_2)(\cos(\varphi_1+\varphi_2) + i\sin(\varphi_1+\varphi_2))$$
So to multiply two complex numbers in polar coordinates, you multiply their lengths and add their angles. I personally think this is incredibly helpful for visualization, and this also shows why the imaginary axis needs to be at a right angle to the real axis: since the angle of $-1$ is $180^\circ$, the angle of $i$ needs to be $90^\circ$ or $270^\circ$.
Best Answer
The line perpendicular to the line $z \bar{a} + \bar{z}a + b =0$ will be given by equation $(i z) \bar{a} + (\overline{iz})a + c =0$ with arbitrary $c\in\mathbb{R}$, which, using the fact that $\bar{i}=-i$, you can write as $$ i(z \bar{a} - \bar{z}a) + c =0 $$ $$ z \bar{a} - \bar{z}a -i c =0 $$