The line $\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-1}{3}$ and the plane $x+2y+z=6$ meet at

Direction ratios. of given line are $(1, -2, 3 )$and the d.r. of normal to the given plane are $(1,2,1)$

Since $1\times 1+(-2)\times 2+3\times 1=0$, therefore, the line is parallel to the plane, 

Also, the base point of the line ($1, 2, 1$) lies in the given plane.

($1+2\times 2+1=6$ is true)

Hence, the given line lies in the given plane.