Let the line L_1 : x + 3 = 0 intersect the lines L_2 : x - y = 0...

The intersection point of $L_1: x = -3$ and $L_2: x - y = 0$ is $A(-3, -3)$.

The intersection point of $L_1: x = -3$ and $L_3: 3x + y = 0$ is $B(-3, 9)$.

The lines $L_2$ and $L_3$ intersect at the origin $O(0, 0)$.

The distances from the origin to $A$ and $B$ are:
$OA = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}$
$OB = \sqrt{(-3)^2 + 9^2} = 3\sqrt{10}$

The angle $\angle AOB$ between the segments $OA$ and $OB$ can be determined using the dot product of vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$:
$\overrightarrow{OA} \cdot \overrightarrow{OB} = (-3)(-3) + (-3)(9) = 9 - 27 = -18$

Since the dot product is negative, $\cos(\angle AOB) < 0$, which means $\angle AOB$ is an obtuse angle. Therefore, the internal bisector of $\angle AOB$ is the bisector of the obtuse angle between the lines $L_2$ and $L_3$.

According to the internal angle bisector theorem in $\triangle AOB$, the bisector of $\angle AOB$ divides the opposite side $AB$ at point $C$ in the ratio of the adjacent sides:
$\dfrac{AC}{BC} = \dfrac{OA}{OB} = \dfrac{3\sqrt{2}}{3\sqrt{10}} = \dfrac{1}{\sqrt{5}}$

Squaring both sides yields:
$\dfrac{AC^2}{BC^2} = \dfrac{1}{5} \Rightarrow \dfrac{BC^2}{AC^2} = 5$

Thus, $BC^2 : AC^2 = 5 : 1$.

Answer: $5:1$