The plane $x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)=0$ passes through the intersection of :

$x-2 y-3 z=0 ; x-2 y-z=0$ $x+2 y+3 z=0 ; 2 x+y-z=0$ $x+2 y+3 z-4=0 ; 2 x+y-z-5=0$ $2 x+y+z-4=0 ; 2 x+y-z-5=0$

$x+2 y+3 z=0 ; 2 x+y-z=0$

$x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)=0$ $= (x+2 y+3 z)+\lambda(2 x+y-z)=0+0 \lambda$ for line $P_1=d_1$ and $P_2=d_2$ their intersection is $P_1+\lambda P_2 = d_1+\lambda d_2$ so $(x+2 y+3 z)+\lambda(2 x+y-z)=0+0 \lambda$ $P_1:  x+2 y+3 z=0$ $P_2: 2 x+y-z=0$