If a line along a chord of the circle $$4x^2 + 4y^2 + 120x + 675 = 0$$, passes through the point $$(-30, 0)$$ and is tangent to the parabola $$y^2 = 30x$$, then the length of this chord is:

First we rewrite the circle equation $$4x^2 + 4y^2 + 120x + 675 = 0$$ in its standard form. Dividing every term by $$4$$ gives $$x^2 + y^2 + 30x + \tfrac{675}{4} = 0.$$ To complete the square for the $$x$$-terms we use $$(x + 15)^2 = x^2 + 30x + 225.$$ Substituting this, we obtain $$(x + 15)^2 - 225 + y^2 + \tfrac{675}{4} = 0.$$ Combining the constants, $$-225 + \tfrac{675}{4} = -\tfrac{225}{1} + \tfrac{675}{4} = -\tfrac{900}{4} + \tfrac{675}{4} = -\tfrac{225}{4}.$$ Hence $$(x + 15)^2 + y^2 = \tfrac{225}{4}.$$ So the centre of the circle is $$(-15,\,0)$$ and the radius is $$R = \sqrt{\tfrac{225}{4}} = \tfrac{15}{2}.$$

The parabola is $$y^2 = 30x,$$ which can be compared with $$y^2 = 4ax.$$ Here $$4a = 30 \;\Rightarrow\; a = 7.5.$$

For the parabola $$y^2 = 4ax,$$ the standard tangent at a point $$(x_1,\,y_1)$$ on it is $$y\,y_1 = 2a\,(x + x_1).$$ Using $$2a = 15,$$ the tangent becomes $$y\,y_1 = 15\,(x + x_1).$$

The required tangent must also pass through the given point $$(-30,\,0).$$ Substituting $$x = -30,\; y = 0$$ into the tangent equation gives $$0 \cdot y_1 = 15\,(-30 + x_1) \;\Longrightarrow\; -30 + x_1 = 0 \;\Longrightarrow\; x_1 = 30.$$

Since $$(x_1,\,y_1)$$ lies on the parabola, we have $$y_1^2 = 30x_1 = 30 \times 30 = 900,$$ so $$y_1 = \pm 30.$$ Thus there are two tangents through $$(-30,0):$$

1. When $$y_1 = 30:$$ $$30y = 15(x + 30) \;\Rightarrow\; 2y = x + 30 \;\Rightarrow\; x - 2y + 30 = 0.$$

2. When $$y_1 = -30:$$ $$-30y = 15(x + 30) \;\Rightarrow\; -2y = x + 30 \;\Rightarrow\; x + 2y + 30 = 0.$$

Each of these lines cuts the circle in a chord. For a circle, the length of the chord cut by a line is given by $$\text{Chord length} = 2\sqrt{R^2 - d^2},$$ where $$d$$ is the perpendicular distance from the centre to the line.

We calculate this distance for one of the lines; both will give the same value because they are symmetric.

Taking $$x - 2y + 30 = 0,$$ the distance from the centre $$(-15,0)$$ is $$d = \frac{|\,(-15) - 2(0) + 30\,|}{\sqrt{1^2 + (-2)^2}} = \frac{|15|}{\sqrt{1 + 4}} = \frac{15}{\sqrt{5}}.$$

Now we evaluate $$R^2 - d^2 = \left(\tfrac{15}{2}\right)^2 - \left(\tfrac{15}{\sqrt{5}}\right)^2 = \tfrac{225}{4} - \tfrac{225}{5} = \tfrac{225}{4} - \tfrac{180}{4} = \tfrac{45}{4}.$$

Therefore the chord length is $$2\sqrt{\tfrac{45}{4}} = 2 \times \frac{\sqrt{45}}{2} = \sqrt{45} = 3\sqrt{5}.$$

Hence, the correct answer is Option C.