A plane which containing the line

$x+2y+3z+1=0=x-y-z-6$ is

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

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

This plane is perpendicular to the plane $-2x+y+z+8=0$ then

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

$\Rightarrow \lambda =\frac{3}{4}$

Equation of plane is

$\frac{7}{4}x+\frac{5}{4}y+\frac{9}{4}z=\frac{14}{4}$

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

Clearly, $(0,1,1)$ lies on the plane.