Let a line L pass through the point of intersection of the lines $$bx + 10y - 8 = 0$$ and $$2x - 3y = 0$$, $$b \in \mathbb{R} - \{\frac{4}{3}\}$$. If the line L also passes through the point (1, 1) and touches the circle $$17(x^2 + y^2) = 16$$, then the eccentricity of the ellipse $$\frac{x^2}{5} + \frac{y^2}{b^2} = 1$$ is

The lines $$bx + 10y - 8 = 0$$ and $$2x - 3y = 0$$ intersect where $$x = \dfrac{3y}{2}$$, giving $$\dfrac{3by}{2} + 10y = 8$$, so $$y = \dfrac{16}{3b + 20}$$ and $$x = \dfrac{24}{3b + 20}$$.

Line $$L$$ passes through this intersection point and through $$(1, 1)$$. The equation of $$L$$ can be written as a member of the family:

$$(bx + 10y - 8) + \lambda(2x - 3y) = 0$$

Since $$L$$ passes through $$(1, 1)$$: $$(b + 10 - 8) + \lambda(2 - 3) = 0$$, so $$b + 2 - \lambda = 0$$, giving $$\lambda = b + 2$$.

The equation of $$L$$ becomes: $$(b + 2(b+2))x + (10 - 3(b+2))y - 8 = 0$$

$$= (3b + 4)x + (4 - 3b)y - 8 = 0$$

Line $$L$$ touches the circle $$17(x^2 + y^2) = 16$$, i.e., $$x^2 + y^2 = \dfrac{16}{17}$$. The distance from the center $$(0, 0)$$ to the line equals the radius $$\dfrac{4}{\sqrt{17}}$$.

$$\dfrac{|0 + 0 - 8|}{\sqrt{(3b+4)^2 + (4-3b)^2}} = \dfrac{4}{\sqrt{17}}$$

$$\dfrac{64}{(3b+4)^2 + (4-3b)^2} = \dfrac{16}{17}$$

$$(3b+4)^2 + (4-3b)^2 = 64 \cdot \dfrac{17}{16} = 68$$

Expanding: $$9b^2 + 24b + 16 + 16 - 24b + 9b^2 = 68$$

$$18b^2 + 32 = 68$$, so $$18b^2 = 36$$, giving $$b^2 = 2$$.

The ellipse is $$\dfrac{x^2}{5} + \dfrac{y^2}{b^2} = 1$$ with $$b^2 = 2$$. Since $$b^2 < 5$$, the semi-major axis is along the $$x$$-axis with $$a^2 = 5$$.

The eccentricity is $$e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{2}{5}} = \sqrt{\dfrac{3}{5}} = \dfrac{\sqrt{3}}{\sqrt{5}}$$.

The correct answer is Option B: $$\dfrac{\sqrt{3}}{\sqrt{5}}$$.