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

D.R of given line are 1, –2, 3 and the d. r. of normal of the given plane are 1, 2, 1.

Since$1\times 1+\left(-2\right)\times 2+3\times 1=0$, therefore, the line is parallel to the plane, also, the base point of the line (1, 2, 1) lies in the given plane. $\left(1+2\times 2+1=6istrue\right)$

Hence, the given line in the given plane.

Alternatively, any point on the given line is

$\left(t+1,-2t+2,3t+1\right)$.

It lies in the given plane $x+2y+z=6$if $t+1+2\left(-2t+2\right)+3t+1=6$

i.e. if 0t=0 , which is true for all real t . Hence every point on the given line lies in the given plane. The line lies in the plane.