The straight line $x+y+1=0$ bisects an angle between a pair of lines, of which one is $2 x-3 y+4=0$. Then the equation of the other line in that pair is

Point of intersection of lines $x+y+1=0$ and $$ 2 x-3 y+4=0 \text { is } A\left(-\frac{7}{5}, \frac{2}{5}\right) \text {. } $$ Now, Let a point $P(-2,0)$ on the line $2 x-3 y+4=0$ and the image of point $P(-2,0)$ w.r.t. line $x+y+1=0$ lying on the other line in the bisector pair. If the coordinate of image of point $P(-2,0)$ w.r.t. line $x+y+1=0$ is $(h, k)$, then $$ \begin{array}{ll} \frac{h+2}{1}=\frac{k-0}{1}=-2 \times \frac{-2+1}{2}=1 \\ \Rightarrow \quad h=-1 \text { and } k=1 \\ \Rightarrow \quad & (h, k)=(-1,1) \end{array} $$ Now, equation of required bisector of angle is line joining the points $A\left(-\frac{7}{5}, \frac{2}{5}\right)$ and $(-1,1)$ is $$ \begin{gathered} y-1=\frac{\frac{3}{5}}{\frac{2}{5}}(x+1) \Rightarrow 2 y-2=3 x+3 \\ 3 x-2 y+5=0 \end{gathered} $$ Hence, option (d) is correct.