Ratio of Squared Distances on Angle Bisector Intersection

To find the ratio $BC^2 : AC^2$, we determine the points of intersection $A$, $B$, and $C$ on the line $L_1: x + 3 = 0$.

1. Intersecting $L_1$ with $L_2: x - y = 0$ gives $A = (-3, -3)$.
2. Intersecting $L_1$ with $L_3: 3x + y = 0$ gives $B = (-3, 9)$.
3. For the lines $L_2$ and $L_3$, we have $a_1 a_2 + b_1 b_2 = 1(3) + (-1)(1) = 2 > 0$, so the obtuse angle bisector corresponds to the positive sign: $\frac{x - y}{\sqrt{2}} = \frac{3x + y}{\sqrt{10}} \implies \sqrt{5}(x - y) = 3x + y$
4. Substituting $x = -3$ into the bisector equation yields $y = -6 + 3\sqrt{5}$, giving $C = (-3, -6 + 3\sqrt{5})$.

Since $A, B, C$ lie on the vertical line $x = -3$, the distances depend only on their $y$-coordinates: $BC = |9 - (-6 + 3\sqrt{5})| = 3(5 - \sqrt{5})$ $AC = |-3 - (-6 + 3\sqrt{5})| = 3(\sqrt{5} - 1)$

Squaring these distances gives: $BC^2 = 90(3 - \sqrt{5})$ $AC^2 = 18(3 - \sqrt{5})$

Thus, $BC^2 : AC^2 = \frac{90(3 - \sqrt{5})}{18(3 - \sqrt{5})} = 5 : 1$.

The correct option is A.