Let $$P : y^2 = 4ax$$, $$a > 0$$ be a parabola with focus $$S$$. Let the tangents to the parabola $$P$$ make an angle of $$\frac{\pi}{4}$$ with the line $$y = 3x + 5$$ touch the parabola $$P$$ at $$A$$ and $$B$$. Then the value of $$a$$ for which $$A$$, $$B$$ and $$S$$ are collinear is:

We are given the parabola $$P: y^2 = 4ax$$, $$a > 0$$, with focus $$S = (a, 0)$$, and we seek the tangents to $$P$$ that make an angle of $$\frac{\pi}{4}$$ with the line $$y = 3x + 5$$, touching $$P$$ at points $$A$$ and $$B$$. The line $$y = 3x + 5$$ has slope 3, so if a tangent has slope $$m$$ and makes an angle of $$\frac{\pi}{4}$$ with this line, then $$\tan\frac{\pi}{4} = \left|\frac{m - 3}{1 + 3m}\right| = 1$$. Case 1: $$\frac{m - 3}{1 + 3m} = 1 \Rightarrow m - 3 = 1 + 3m \Rightarrow -2m = 4 \Rightarrow m = -2$$. Case 2: $$\frac{m - 3}{1 + 3m} = -1 \Rightarrow m - 3 = -1 - 3m \Rightarrow 4m = 2 \Rightarrow m = \frac{1}{2}$$.

For the parabola $$y^2 = 4ax$$, a tangent with slope $$m$$ touches at the point $$\left(\frac{a}{m^2}, \frac{2a}{m}\right)$$, yielding for $$m = -2$$ the point $$A = \left(\frac{a}{4}, -a\right)$$ and for $$m = \frac{1}{2}$$ the point $$B = (4a, 4a)$$.

Three points are collinear if the area of the triangle they form is zero: Area $$= \frac{1}{2}|x_A(y_B - y_S) + x_B(y_S - y_A) + x_S(y_A - y_B)| = \frac{1}{2}\left|\frac{a}{4}(4a - 0) + 4a(0 - (-a)) + a(-a - 4a)\right| = \frac{1}{2}\left|a^2 + 4a^2 - 5a^2\right| = \frac{1}{2}|0| = 0$$, so the area is always zero regardless of the value of $$a$$.

Hence, $$A$$, $$B$$, and $$S$$ are collinear for any $$a > 0$$. Therefore, the answer is Option D: any $$a > 0$$.