1. Convert the given complex numbers to points in the Argand plane

Given: $z_1=3+4i \Rightarrow (3,4), \quad z_2=4+3i \Rightarrow (4,3), \quad z_3=5i \Rightarrow (0,5)$

We first find the circle passing through these three points.

2. Equation of the circle

Let the circle be $x^2+y^2+2gx+2fy+c=0$

Since it passes through $(3,4)$: $3^2+4^2+2g(3)+2f(4)+c=0$ $25+6g+8f+c=0 \qquad (1)$

Through $(4,3)$: $16+9+8g+6f+c=0$ $25+8g+6f+c=0 \qquad (2)$

Through $(0,5)$: $25+10f+c=0 \qquad (3)$

From $(1)-(2)$: $6g+8f-(8g+6f)=0$ $-2g+2f=0 \Rightarrow f=g$

Using (3): $c=-25-10f$

Substitute in (1): $25+6f+8f-25-10f=0$ $4f=0 \Rightarrow f=0$ Thus, $g=0, \quad c=-25$

Hence the circle is $x^2+y^2=25$

So the circle is centered at origin with radius $5$.

3. Direction of the line through $z_2$ and $z_3$

Points are $(4,3)$ and $(0,5)$. Its slope is $m_{23}=\frac{5-3}{0-4}=\frac{2}{-4}=-\frac12$

A line perpendicular to this has slope $m=2$

So the required line through $z_1=(3,4)$ is $y-4=2(x-3)$ $y=2x-2$

Point $z$ lies on this line and on the circle $x^2+y^2=25$.

4. Find the second intersection of the line with the circle

Substitute $y=2x-2$ into the circle: $x^2+(2x-2)^2=25$ $x^2+4x^2-8x+4=25$ $5x^2-8x-21=0$

Solve: $5x^2-8x-21=(x-3)(5x+7)=0$

So, $x=3 \quad \text{or} \quad x=-\frac75$

The point $x=3$ gives $y=4$, which is $z_1$ itself. Thus the other point is $x=-\frac75, \quad y=2\left(-\frac75\right)-2=-\frac{14}{5}-\frac{10}{5}=-\frac{24}{5}$

Hence $z=-\frac75-\frac{24}{5}i$

5. Find the argument of $z$

Now, $z=-\frac75-\frac{24}{5}i$

This lies in the third quadrant. Therefore, $\arg(z)=\tan^{-1}\left(\frac{y}{x}\right)-\pi$ Since both $x,y<0$, $\left|\frac{y}{x}\right|=\frac{24/5}{7/5}=\frac{24}{7}$

Thus, $\arg(z)=\tan^{-1}\left(\frac{24}{7}\right)-\pi$

6. Match with options

This is exactly Option B.

$\boxed{\arg(z)=\tan^{-1}\left(\frac{24}{7}\right)-\pi}$