The projection of the line $\frac{x+1}{-1}=\frac{y}{2}=\frac{z-1}{3}$ on the plane $x-2y+z=6$ is the line of intersection of this plane with the plane 

$\text{ Equation of the plane through }(-1,0,1)\text{ is }a(x+1)+b(y-0)+c(z-1)=0$

Required plane is perpendicular to given plane $\Rightarrow a-2b+c=0,$ is parallel to given line $\Rightarrow -a+2b+3c=0$

On solving we get $c=0,a=2b\Rightarrow \frac{a}{2}=\frac{b}{1}=\frac{c}{0}$

$2b(x+1)+by=0\Rightarrow 2x+y+2=0$