A plane P contains the line

Equation of plane P can be assumed as

$P: x+2 y+3 z+1+\lambda(x-y-z-6)=0$

$\Rightarrow \mathrm{P}:(1+\lambda) \mathrm{x}+(2-\lambda) \mathrm{y}+(3-\lambda) \mathrm{z}+1-6 \lambda=0$

$\Rightarrow \overrightarrow{\mathrm{n}}_{1}=(1+\lambda) \hat{\mathrm{i}}+(2-\lambda) \hat{\mathrm{j}}+(3-\lambda) \hat{\mathrm{k}}$

$\therefore \quad \vec{n}_{1} \cdot \vec{n}_{2}=0$

$\Rightarrow 2(1+\lambda)-(2-\lambda)-(3-\lambda)=0$

$\Rightarrow 2+2 \lambda-2+\lambda-3+\lambda=0 \Rightarrow \lambda=\frac{3}{4}$\

$\Rightarrow P: \frac{7 x}{4}+\frac{5}{4} y+\frac{9 z}{4}-\frac{14}{4}=0$

$\Rightarrow 7 x+5 y+9 z=14$

$(0,1,1)$ lies on $\mathrm{P}$