A plane $\mathrm{P}$ contains the line $x+2 y+3 z+1=0=x-y-z-6$ and is perpendicular to the plane $-2 x+y+z+8=0$.
Then which of the following points lies on P?
Correct Option: , 2
Equation of plane $\mathrm{P}$ can be assumed as
$P: x+2 y+3 z+1+\lambda(x-y-z-6)=0$
$\Rightarrow P:(1+\lambda) x+(2-\lambda) y+(3-\lambda) z+1-6 \lambda=0$
$\Rightarrow \overrightarrow{\mathrm{n}}_{1}=(1+\lambda) \hat{\mathrm{i}}+(2-\lambda) \hat{\mathrm{j}}+(3-\lambda) \hat{\mathrm{k}}$
$\therefore \quad \vec{n}_{1} \cdot \vec{n}_{2}=0$
$\Rightarrow 2(1+\lambda)-(2-\lambda)-(3-\lambda)=0$
$\Rightarrow 2+2 \lambda-2+\lambda-3+\lambda=0 \Rightarrow \lambda=\frac{3}{4}$
$\Rightarrow P: \frac{7 x}{4}+\frac{5}{4} y+\frac{9 z}{4}-\frac{14}{4}=0$
$\Rightarrow 7 x+5 y+9 z=14$
$(0,1,1)$ lies on $\mathrm{P}$