The equation of one of the straight lines which passes through the point $$(1, 3)$$ and makes an angles $$\tan^{-1}(\sqrt{2})$$ with the straight line, $$y + 1 = 3\sqrt{2}x$$ is:

The given line is $$y + 1 = 3\sqrt{2}x$$, so its slope is $$m_1 = 3\sqrt{2}$$. The angle between the required line and this line is $$\theta = \tan^{-1}(\sqrt{2})$$, so $$\tan\theta = \sqrt{2}$$.

Using the angle formula $$\tan\theta = \left|\frac{m - m_1}{1 + mm_1}\right|$$, where $$m$$ is the slope of the required line, we get $$\sqrt{2} = \left|\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m}\right|$$.

Taking the positive case: $$m - 3\sqrt{2} = \sqrt{2}(1 + 3\sqrt{2}m) = \sqrt{2} + 6m$$, giving $$-5m = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2}$$, so $$m = -\frac{4\sqrt{2}}{5}$$.

Taking the negative case: $$m - 3\sqrt{2} = -\sqrt{2}(1 + 3\sqrt{2}m) = -\sqrt{2} - 6m$$, giving $$7m = 3\sqrt{2} - \sqrt{2} = 2\sqrt{2}$$, so $$m = \frac{2\sqrt{2}}{7}$$.

The line through $$(1, 3)$$ with slope $$m = -\frac{4\sqrt{2}}{5}$$ is $$y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$$. Multiplying by 5: $$5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$$, which gives $$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$$.

The answer is Option A: $$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$$.