The plane containing the line $\frac{x+1}{2}=\frac{2y-3}{4}=-z$ and parallel to the line $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z-3}{4}$ has the vector equation

Required plane is passing through  $\overline{a}=\left(-1,\frac{3}{2},0\right)$ and parallel to vectors

$\overline{b}=(2,2,-1),\overline{c}=(3,2,4)$

$\therefore$ $\text{ Equation of the plane is given by }$

$\overline{r}=\overline{a}+t\overline{b}+s\overline{c}\text{ wheret, s are scalars }$