What is $\max(|z|)$ if $\Bigl|\frac{6z - i}{2 + 3iz}\Bigr|\leq1$

Question

I'm solving a multiple choice question but my answer is not among the choices and I don't know where I am wrong. The question is this:

What is the $\max|z|$ if: $$\Bigl|\frac{6z - i}{2 + 3iz}\Bigr|\leq1$$ and $z$ is a complex number.

1. $\frac{1}{5}$
2. $\frac{1}{4}$
3. $\frac{1}{3}$
4. $\frac{1}{2}$

My solution is this but my answer isn't among them:

$$ \biggl|\frac{6z - i}{2 + 3iz}\biggr|\leq1\Rightarrow |6z-i| \leq |2+3iz| \Rightarrow |6z-i|^2 \leq |2+3iz|^2 $$ $$ \Rightarrow (6z-i)\overline{(6z-i)} \leq (2+3iz)\overline{(2+3iz)} $$ $$ \Rightarrow (6z-i)(6\bar{z}+i) \leq (2+3iz)(2-3i\bar{z}) $$ $$ \Rightarrow 36|z|^2 +6iz - 6i\bar{z} + 1 \leq 4 - 6i\bar{z} +6iz + 9|z|^2 $$ $$ \Rightarrow 17|z|^2 \leq 3 \Rightarrow |z|^2 \leq \frac{3}{17} \Rightarrow |z| \leq \sqrt{\frac{3}{17}}$$

Where am I wrong? what is the right solution? Thanks in advance.

Answer 1

$36-9=27$. You have taken it as $17$. The correct answer is 3).