The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines…
The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is
$8 x+y-13 z+27=0$
$2 x+y+z=0$
$3 x-y-z-10=0$
$8 x+14 y+13 z+37=0$
Solution
Equation of the plane passing through $\overline{\mathrm{a}}$ and parallel to $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is $\overline{\mathrm{r}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$
$\begin{aligned} & \therefore \quad \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 2 & -4 \\ 2 & -3 & 2\end{array}\right|=-8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\ & \begin{aligned} \therefore & \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =(2)(-8)+(-1)(-14)+(-3)(-13) \\ & =-16+14+39=37\end{aligned}\end{aligned}$
$\therefore \quad$ Required equation is
$\bar{r} \cdot(-8 \hat{\mathbf{i}}-14 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})=37$
i.e., $8 x+14 y+13 z+37=0$