Equation of ${\mathrm{P}}_3$ is ${\mathrm{P}}_1+\mathrm{\lambda }{\mathrm{P}}_2=0$
$\mathrm{x}\left(1+2\mathrm{\lambda }\right)+\mathrm{y}\left(1-\mathrm{\lambda }\right)+\mathrm{z}\left(2+3\mathrm{\lambda }\right)-4+5\mathrm{\lambda }=0$
$\overrightarrow{\mathrm{A}\mathrm{B}}=<2,-\mathrm{1,3}>$
$\mathrm{a}$ normal vector to plane ${\mathrm{P}}_3$ is $<1+2\mathrm{\lambda },\mathrm{}1-\mathrm{\lambda },\mathrm{}2+3\mathrm{\lambda }>$
If $\mathrm{A}\mathrm{B}$ is parallel to ${\mathrm{P}}_3=0$
$\Rightarrow 2\left(1+2\mathrm{\lambda }\right)-1\left(1-\mathrm{\lambda }\right)+3\left(2+3\mathrm{\lambda }\right)=0$
$\Rightarrow 2+4\mathrm{\lambda }-1+\mathrm{\lambda }+6+9\mathrm{\lambda }=0$
$\Rightarrow 14\mathrm{\lambda }+7=0$
$\Rightarrow \mathrm{\lambda }=-\frac{1}{2}$
Equation of required plane is $0+\mathrm{y}\times \frac{3}{2}+\mathrm{z}\left(\frac{1}{2}\right)-\frac{13}{2}=0$

i.e. $3\mathrm{y}+\mathrm{z}-13=0$