Let a circle $$C$$ in complex plane pass through the points $$z_1 = 3 + 4i, z_2 = 4 + 3i$$ and $$z_3 = 5i$$. If $$z \neq z_1$$ is a point on $$C$$ such that the line through $$z$$ and $$z_1$$ is perpendicular to the line through $$z_2$$ and $$z_3$$, then $$\arg z$$ is equal to

We need to find $$\arg z$$ for a point $$z \neq z_1$$ on the circle $$C$$ such that the line through $$z$$ and $$z_1$$ is perpendicular to the line through $$z_2$$ and $$z_3$$. The circle passes through $$z_1 = 3 + 4i$$, $$z_2 = 4 + 3i$$, and $$z_3 = 5i$$, corresponding to the points $$(3,4)$$, $$(4,3)$$, and $$(0,5)$$.

Using the general equation $$x^2 + y^2 + Dx + Ey + F = 0$$ leads to three linear equations: from $$(3,4)$$ we get $$3D + 4E + F = -25$$, from $$(4,3)$$ we have $$4D + 3E + F = -25$$, and from $$(0,5)$$ we deduce $$5E + F = -25$$. Subtracting the second equation from the first gives $$-D + E = 0$$, so $$D = E$$, and then $$F = -25 - 5E$$. Substituting back into the first equation leads to $$3E + 4E + (-25 - 5E) = -25$$, which simplifies to $$2E = 0$$, hence $$E = 0$$, followed by $$D = 0$$ and $$F = -25$$. Therefore, the circle is $$x^2 + y^2 = 25$$.

Next, the slope of the line through $$z_2$$ and $$z_3$$ is determined by the direction $$z_3 - z_2 = 5i - (4+3i) = -4 + 2i$$, which gives a slope of $$\frac{2}{-4} = -\frac{1}{2}$$. A perpendicular slope is therefore $$2$$, so the line through $$(3,4)$$ can be written as $$y - 4 = 2(x - 3)$$, or equivalently $$y = 2x - 2$$.

To find the intersection of this line with the circle, we substitute $$y = 2x - 2$$ into $$x^2 + y^2 = 25$$, which leads to
$$x^2 + (2x-2)^2 = 25$$
$$x^2 + 4x^2 - 8x + 4 = 25$$
$$5x^2 - 8x - 21 = 0$$. The quadratic formula then gives
$$x = \frac{8 \pm \sqrt{64 + 420}}{10} = \frac{8 \pm \sqrt{484}}{10} = \frac{8 \pm 22}{10}$$, hence $$x = 3$$ (which corresponds to $$z_1$$) or $$x = -\frac{7}{5}$$, and in the latter case $$y = 2\left(-\frac{7}{5}\right) - 2 = -\frac{24}{5}$$.

Therefore, the required point is $$z = -\frac{7}{5} - \frac{24}{5}i$$, which lies in the third quadrant. Since $$\tan\theta = \frac{-24/5}{-7/5} = \frac{24}{7}$$ and the point is in the third quadrant, its argument is given by
$$\arg z = \tan^{-1}\frac{24}{7} - \pi$$.

The correct answer is Option A: $$\tan^{-1}\dfrac{24}{7} - \pi$$.