Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
  \begin{array}{l}
    2x\equiv 3\pmod 5 \\
    4x\equiv 2\pmod 6 \\
    3x\equiv 2\pmod 7
  \end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, giving
$N_1 = \frac{210}{5} = 42, \enspace N_2 = \frac{210}{6} = 35, \enspace N_3 = \frac{210}{7} = 30 $.
$$
\begin{align}
42z_1&\equiv 1\pmod 5\Rightarrow\enspace\enspace\;2z_1\equiv 1\pmod 5\Rightarrow\enspace &&\overline{z_1}=\overline{3}\\
35z_2&\equiv 1\pmod 6\Rightarrow\enspace-1z_2\equiv 1\pmod 6\Rightarrow\enspace &&\overline{z_2}=\overline{-1}\\
30z_3&\equiv 1\pmod 7\Rightarrow\enspace\enspace\;2z_3\equiv 1\pmod 7\Rightarrow\enspace &&\overline{z_1}=\overline{4}
\end{align}
$$
So the solution should be
$$
\begin{align}
\overline{x} &= \overline{3\times42\times3} + \overline{2\times35\times(-1)}+\overline{2\times30\times4}\\ 
&= \overline{378-70+240}\\
&= \overline{548}\\
&= \overline{128}
\end{align}
$$
Which is clearly wrong, so I'm wondering which additional steps I need to take to get to the correct answer.
Thanks in advance.

You seem to have solved $$\left\{\begin{array}{c} x\equiv3 \pmod{5}\\x\equiv 2\pmod{6}\\x\equiv 2 \pmod{7}\end{array}\right.$$
But you need to take into account the coefficients on $x$ in each congruence.
$2x\equiv 3\pmod{5}$ means $x\equiv 4\pmod{5}$
$4x\equiv 2\pmod{6}$ means $x\equiv 2 \mbox{ or } 5 \pmod{6}$
$3x\equiv 2\pmod{7}$ means $x\equiv 3\pmod{7}$
