You can't just choose a point on the line. The point on the line closest to (0, 0, 1) is at the foot of the line perpendicular to the given line through (0, 0, 1). Of course, the perpendicular line will lie on a plane perpendicular to the given line.
The given line, (x-1)/2 = (y-2) /4 = (z-3) /4 can also be written in parametric equations by letting the parameter, t, be equal to each of those: t= (x-1)/2 so x= 1+ 2t. t= (y-2)/4 so y= 2+ 4t. t= (z-3)/4 so z= 3+ 4t.
Since the given line has "direction vector" <2, 4, 4>, any plane perpendicular to the line has equation 2x+ 4y+ 4z= C for some constant C. When x= 0, y= 0, z= 1, 2(0)+ 4(0)+ 4(1)= 4 so the equation of the plane perpendicular to the given line and containing (0, 0, 1) is 2x+ 4y+ 4z= 4.
Find the point where the given line crosses that plane. The distance from (0, 0, 1) to the point you just found is the distance from the point to the line.