The plane $x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)=0$ passes through the intersection of :
Answer & explanation
Correct answer: option 2
$x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)=0$
$= (x+2 y+3 z)+\lambda(2 x+y-z)=0+0 \lambda$
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 $(x+2 y+3 z)+\lambda(2 x+y-z)=0+0 \lambda$
$P_1: x+2 y+3 z=0$
$P_2: 2 x+y-z=0$