Let $z_1, z_2$ lie on $|z-5i|=3$. A tangent is drawn at $z_2$, cutting the real axis at $z_3$. Show that $\operatorname{Arg}(5i-z_3/z_2+z_3) = \pi/6$. A circle $|z-5i|=3$ has two points $z_1$ and $z_2$ on it such that $\lvert z_1\rvert < \lvert z_2 \rvert$, $\arg(z_1) = \arg(z_2) = \pi/3$. A tangent is drawn at $z_2$, cutting the real axis at $z_3$. We need to prove that $\operatorname{Arg}(5i-z_3/z_2+z_3) = \pi/6$.
I tried to solve this question by forming equations for the circle and the tangent and then solving them. I got the value of $z_2$ as $(5\sqrt{3}+\sqrt{11})/4 + i\sqrt{3}(5\sqrt{3}+\sqrt{11})/4$ and $z_3$ as $\sqrt{11}$, but when I am putting these values in the expression $\operatorname{Arg}(5i-z_3/z_2+z_3)$, the argument doesn't come out to be $\pi/6$. Someone please help me correct my mistake here.
Also, I'll be really grateful if someone suggests a better approach to this question for I find my method quite lengthy, and based more on conic sections (circles), rather than complex numbers.
Edit:
A shorter solution exists on Byju's website but not able to understand how they took angle ADB to be $60$, where A is z3, B is z2 and D is centre of circle.
 A: You correctly got
$$z_2=\frac{5\sqrt{3}+\sqrt{11}}{4} + i\sqrt{3}\cdot \frac{5\sqrt{3}+\sqrt{11}}{4},\qquad z_3=\sqrt{11}$$
but you cannot prove that $\operatorname{Arg}(5i-z_3/z_2+z_3) = \pi/6$ because it is false that $\operatorname{Arg}\bigg(5i-\dfrac{z_3}{z_2}+z_3\bigg) = \pi/6$. Also, it is false that $\operatorname{Arg}\bigg(\dfrac{5i-z_3}{z_2+z_3}\bigg)=\pi/6$.

If what we want to prove is $$\operatorname{Arg}\bigg(\frac{5i-z_3}{z_2\color{red}-z_3}\bigg) = \frac{\pi}6\tag1$$
which is true, then there is a simple solution. You don't need to find $z_2$.

Edit:
A shorter solution exists on Byju's website but not able to understand how they took angle ADB to be $60$, where A is z3, B is z2 and D is centre of circle.

Since $\angle{AOD}=\angle{ABD}=90^\circ$ where $O$ is the origin, the four points $A,B,D,O$ are on the circle whose diameter is $AD$.
So we get, from the inscribed angle theorem,
$$\angle{ADB}=\angle{AOB}=60^\circ$$
from which we get $(1)$ because
$$\operatorname{Arg}\bigg(\frac{5i-z_3}{z_2-z_3}\bigg) =\angle{BAD}$$
