The given plane has normal vector
[tex]x-2y=5\implies\mathbf n=\langle1,-2,0\rangle[/tex]
Scaling n by a real number t gives a set of vectors that span an entire line through the origin. Translating this line by adding the vector <2, 1, 1> makes it so that this line passes through the point (2, 1, 1). So this line has equation
[tex]\mathbf r(t)=\langle1,-2,0\rangle t+\langle2,1,1\rangle=\langle 2+t, 1-2t, 1\rangle[/tex]
This line passes through (2, 1, 1) when t = 0, and the line intersects with the plane when
[tex]x-2y=5\implies(2+t)-2(1-2t)=5\implies5t=5\implies t=1[/tex]
which corresponds the point (3, -1, 1) (simply plug t = 1 into the coordinates of [tex]\mathbf r(t)[/tex]).
So the distance between the plane and the point is the distance between the points (2, 1, 1) and (3, -1, 1):
[tex]\sqrt{(2-3)^2+(1-(-1))^2+(1-1)^2}=\boxed{\sqrt5}[/tex]
