Let the angle
bisector of
$$
\begin{aligned}
& 3 x+2 y+4 \text { and } \\
& y=m x+c \text { is } \\
& \quad 2 x+3 y+1=0
\end{aligned}
$$
Slope of line
$$
\begin{aligned}
& 3 x+2 y+4 \text { is } \\
& \qquad m_1=\frac{-3}{2}
\end{aligned}
$$


Slope of line $2 x+3 y+1=0$ is $m_2=\frac{-2}{3}$
Angle between lines and bisector are equal,
$$
\begin{aligned}
& \text { so }\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{m-m_2}{1+m_1 m_2}\right| \\
& \Rightarrow \frac{-2 / 3+3 / 2}{1+1}=\frac{m+2 / 3}{1-\frac{2 m}{3}} \Rightarrow \frac{5}{12}=\frac{m+2 / 3}{1-2 m / 3} \\
& \Rightarrow \quad 5-\frac{10 m}{3}=12 m+8 \Rightarrow\left(\frac{10}{3}+12\right) m=-3 \\
& \Rightarrow \quad \frac{46}{3} m=-3 \Rightarrow m=-\frac{9}{46}
\end{aligned}
$$
So, equation of required line is

Now, for intersection point, equation $3 x+2 y+4=0$ and $2 x+3 y+1=0$, is given by
$$
\begin{aligned}
6 x+4 y+8 & =0 \\
6 x+9 y+3 & =0 \\
-\quad-\quad & \\
\hline 5 y & =5 \\
y & =1 \\
x & =-2
\end{aligned}
$$
and
intersection point is $(-2,1)$.
Required line is also passes through $(-2,1)$
$$
\text { So, } 1=\frac{-9}{46}(-2)+c \Rightarrow c=1-\frac{9}{23} \Rightarrow c=\frac{14}{23}
$$
Put in Eq. (i), we get
$$
\begin{aligned}
y & =\frac{-9}{46} x+\frac{14}{23} \\
\Rightarrow \quad 46 y & =-9 x+28 \Rightarrow 9 x+46 y-28=0
\end{aligned}
$$