Any complex number can be written as $z=x+iy$ where $x,y \in \mathbb{R}$ and $i$ is the imaginary unit meaning $i^2=-1$ by definition. Besides the conjugate of a complex number $z$ is denoted by $\bar{z}$ where $\bar{z}=x-iy$.
Now
$\begin{align} (3+7i)z+(10−2i)\bar{z}+100&=0 \\ \Rightarrow (3+7i)(x+iy)+(10−2i)(x-iy)+100&=0 \\ \Rightarrow 3x+3iy+7ix-7y+10x-10iy-2ix-2y+100&=0; ~[\because i^2=-1] \\ \Rightarrow 13x-9y+100+i\left(5x-7y \right)&=0\end{align}$
As $a+ib=0\Leftrightarrow a=0 \mathrm{~and~}b=0$,
$\therefore 13x-9y+100=0$ and $5x-7y=0$ both of which represent two straight lines. They intersect each other because their slopes are different $[\frac{13}{9} \ne \frac{5}{7}]$ meaning they are not parallel. Since they intersect each other, the intersecting point will be the solution to the original equation.
$\therefore (3+7i)z+(10−2i)\bar{z}+100=0~$ represents a point in the complex plane.
So the correct answer is C.