Find the distance between the straight line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{1}$ and the plane $x-y+z-5=0$.

1. Check if the line is parallel to the plane:Line direction vector $\vec{d}=(1,2,1)$.Plane normal vector $\vec{n}=(1,-1,1)$.$\vec{d}\cdot \vec{n}=(1)(1)+(2)(-1)+(1)(1)=1-2+1=0$.Since the dot product is zero, the line is parallel to the plane.2. Calculate distance from a point on the line to the plane:A point on the line is $P(3,4,5)$.Distance to plane $Ax+By+Cz+D=0$:
$$d=\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$$
$$d=\frac{|(1)(3)+(-1)(4)+(1)(5)-5|}{\sqrt{1^2+(-1{)}^2+1^2}}=\frac{|3-4+5-5|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$$