Given that, $|x^{2}-4x-13| = r\quad \longrightarrow (1)$
We know that, $|x| = \left\{\begin{matrix}x \;;&x \geq 0 \\ -x\;; &x<0 \end{matrix}\right.$
$\textbf{Case 1:}\; x^{2}-4x-13 = r$
$\Rightarrow x^{2}-4x-13-r = 0$
For two distinct real roots, the discriminant will be greater than zero.
$\Rightarrow D_{1}>0$
$\Rightarrow b^{2}-4ac>0 \quad [\because \text{For the quadratic equation}\; ax^{2} + bx +c = 0]$
$\Rightarrow (-4)^{2}-4(-13-r)>0$
$\Rightarrow 16+52+4r>0$
$\Rightarrow 4r+68>0$
$\Rightarrow \boxed{r>-17}$
$\textbf{Case 2:}\; x^{2}-4x-13 = -r$
$\Rightarrow x^{2}-4x-13+r = 0$
For one real root, the discriminant will be zero.
$\Rightarrow D_{2} =0$
$\Rightarrow b^{2}-4ac=0$
$\Rightarrow (-4)^{2}-4(1)(-13+r)=0$
$\Rightarrow 16+52-4r=0$
$\Rightarrow 68-4r=0$
$\Rightarrow 4r=68$
$\Rightarrow \boxed{r=17}$
$\therefore$ The value of $r = 17$
Correct Answer $:\text{C}$
$\textbf{PS:}$
We know the quadratic formula is
$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
for any quadratic equation written in standard form of $ax^2+bx+c=0.$ The discriminant $D$ for the quadratic equation is
$$D=b^2-4ac,$$
where
$$\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}$$