Let slope of required line $=m$
The angle between bisector and new line will be same as between bisector $x+y+1$ and line $2 x+3 y-4=0$
$\therefore$ Angle between bisector $x+y+1=0$ and new line is
$$
\tan \theta=\left|\frac{m+1}{1-m}\right|
$$
$\left[\because\right.$ Angle between two lines $\left.\tan \theta=\left|\frac{m_2-m_1}{\mid 1+m_1 m_2}\right|\right]$
And angle between bisector $x+y+1$ and $2 x+3 y-4=0$
$$
\begin{aligned}
& \tan \theta=\left|\frac{\frac{-2}{3}+\frac{1}{3}}{1+\frac{2}{3} \times 1}\right|=\left|\frac{\frac{1}{3}}{\frac{3+2}{3}}\right| \\
& \tan \theta=\left|\frac{1}{5}\right|
\end{aligned}
$$

From Eqs. (i) and (ii),
$$
\begin{aligned}
& \frac{m+1}{1-m}=\left(\frac{1}{5}\right) \text { and } \frac{m+1}{1-m}=\frac{-1}{5} \\
& \Rightarrow 5 m+5=1-m \text { or } 5 m+5=-1+m \\
& \Rightarrow 6 m=-4 \text { or } 4 m=-6
\end{aligned}
$$


$\Rightarrow m=\frac{-2}{3}$ or $m=\frac{-3}{2}$
So, new line has slope of $\frac{-3}{2}$.
Now, for point of intersection of line and bisector, we solve,
$$
\text { and } \quad \begin{aligned}
x+y & =-1 \\
\Rightarrow \quad 2 x+3 y & =4 \\
2 x+2 y & =-2 \\
2 x \pm 3 y & =4 \\
\hline-y & =-6 \\
y & =6
\end{aligned}
$$

Put in Eq. (iii) $x+y=-1$
$$
\Rightarrow \quad x+6=-1 \Rightarrow x=-7
$$

So, coordinate of point of intersection is $(-7,6)$ equation of line passes through $(-7,6)$ and slope $\frac{-3}{2}$
$$
\begin{aligned}
y-y_1=m\left(x-x_1\right) & \Rightarrow(y-6)=\frac{-3}{2}(x+7) \\
\Rightarrow 2 y-12=-3 x-21 & \Rightarrow 3 x+2 y+9=0
\end{aligned}
$$