Equation of a line parallel to the line $z\bar{a} + \bar{z} a + b = 0$ is $z\bar{a} + \bar{z} a + c = 0$, (where c is a real number)
Equation of a line perpendicular to the line $z\bar{a} + \bar{z} a + b = 0$ is $z\bar{a} + \bar{z} a + c\imath= 0$, (where c is a real number)
How is the second one possible when that constant part should be a real number in an equation of a line.
I'm just starting with geometry of complex numbers, please use less rigorous things in your answers.
On
I don't fully understand your way if explaining the problem, but I think you're unfamiliar with the concept that multiplying on the complex plane means rotating (or stretching), and in particular multiplying by $i$ means rotating by $\frac{\pi} {2}$, aka 90°, aka right angle, aka "perpendicular to". You can easily check it by taking a random complex number (and in particular a real number for your problem) and multiplying by $i$. This is why, while on $\mathbb{R} ^2$ we have to do all the fancy stuff to get a perpendicular line to one given, on $\mathbb{C} $ we just multiply by $i$. 3b1b made a very interesting video about this, you should check it out.
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 $$ i(z \bar{a} - \bar{z}a) + c =0 $$ $$ z \bar{a} - \bar{z}a -i c =0 $$