The relation between the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ and the plane$ax+by+cz+d=0$  is 

1. Line is parallel to the plane if  $al+bm+cn=0$and$a{x}_1+b{y}_1+c{z}_1+d\ne 0$
2. Line lies on the plane if  $al+bm+cn=0$and $a{x}_1+b{y}_1+c{z}_1+d=0$
3. Line intersect the plane if $al+bm+cn\ne 0$

Here the given line is $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$  and the plane is $x-2y+z=0$

And  $al+bm+cn=1\left(1\right)-2\left(2\right)+1\left(3\right)\ne 0$

Hence, the line intersects the plane. 

The line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$  also passing through the origin. Hence the equation also can be written as $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$