Let the line $$L_1: x + 3 = 0$$ intersect the lines $$L_2: x - y = 0$$ and $$L_3: 3x + y = 0$$ at the points $$A$$ and $$B$$, respectively. Let the bisector of the obtuse angle between the lines $$L_2$$ and $$L_3$$ intersect the line $$L_1$$ at the point $$C$$. Then $$BC^2 : AC^2$$ is equal to :

$$A = (-3, -3)$$ and $$B = (-3, 9)$$

The lines are $$x - y = 0$$ and $$3x + y = 0$$. Their angle bisectors are given by:

$$\frac{x - y}{\sqrt{1^2 + (-1)^2}} = \pm \frac{3x + y}{\sqrt{3^2 + 1^2}}$$

$$\frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}} \implies \sqrt{5}(x - y) = \pm (3x + y)$$

$$a_1a_2 + b_1b_2 = (1)(3) + (-1)(1) = 3 - 1 = 2 > 0$$

Since it is positive, the $$+$$ sign gives the obtuse angle bisector: $$\sqrt{5}x - \sqrt{5}y = 3x + y \implies (\sqrt{5} - 3)x = (\sqrt{5} + 1)y$$

$$(\sqrt{5} - 3)(-3) = (\sqrt{5} + 1)y_c \implies y_c = \frac{-3(\sqrt{5} - 3)}{\sqrt{5} + 1}$$

$$y_c = \frac{-3(\sqrt{5} - 3)(\sqrt{5} - 1)}{5 - 1} = \frac{-3(5 - 4\sqrt{5} + 3)}{4} = \frac{-3(8 - 4\sqrt{5})}{4} = -6 + 3\sqrt{5}$$

$$\therefore C = (-3, -6 + 3\sqrt{5})$$

$$AC^2 = (y_c - y_a)^2 = (-6 + 3\sqrt{5} - (-3))^2 = (3\sqrt{5} - 3)^2 = 9(\sqrt{5} - 1)^2$$

$$BC^2 = (y_c - y_b)^2 = (-6 + 3\sqrt{5} - 9)^2 = (3\sqrt{5} - 15)^2 = 9(\sqrt{5} - 5)^2 = 9 \cdot 5(1 - \sqrt{5})^2$$

$$\frac{BC^2}{AC^2} = \frac{45(\sqrt{5} - 1)^2}{9(\sqrt{5} - 1)^2} = 5$$