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

$\text{ Given line }\frac{x+1}{-1}=\frac{y}{2}=\frac{z-1}{3}$

$\text{ And plane is }x-2y+z=6$

$\text{ Equation of the plane through }$

$(-1,0,1)\text{ is }a(x+1)+b(y-0)+c(z-1)=0\dots \dots .\left(1\right)$

$\text{ Which is parallel to the given line and perpendicular to the given plane }$

$\Rightarrow -a+2 b+3 c=0$

$\text{ and }a-2b+c=0$

$\Rightarrow c=0,a=2b$

$\text{∴ From (1) equation of the required plane is }2x+y+2=0$