Calculate the distance between the straight line and the plane:
$$r:(x,y,z)=(2,1,0)+k\cdot(1,4-3)$$
$$\pi:x+y+2z-1=0$$
See development and solution
Development:
Let's consider the governing vector of the straight line, $$\vec{v}=(1,4,-3)$$, and the normal vector of the plane, $$\vec{n}=(1,1,2)$$ and we do the scalar product: $$$(1, 4, -3)\cdot(1, 1, 2) = -1 \neq 0$$$ Therefore the straight line and the plane are not parallel and $$\text{d}(r,\pi)=0$$.
Solution:
$$\text{d}(r,\pi) = 0$$