The distance between the line $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+4\hat{k})$ and the plane $\vec{r}\cdot (\hat{i}+5\hat{j}+\hat{k})=5$ is:

The direction of the line is $\vec{d}=(1,-1,4)$.The normal to the plane is $\vec{n}=(1,5,1)$.Check if the line is parallel to the plane:$\vec{d}\cdot \vec{n}=(1)(1)+(-1)(5)+(4)(1)=1-5+4=0$.The line is parallel to the plane. The distance is the perpendicular distance of any point on the line (e.g., $(2,-2,3)$) to the plane $x+5y+z-5=0$.
$$d=\frac{|(1)(2)+(5)(-2)+(1)(3)-5|}{\sqrt{1^2+5^2+1^2}}$$
$$d=\frac{|2-10+3-5|}{\sqrt{27}}=\frac{|-10|}{3\sqrt{3}}=\frac{10}{3\sqrt{3}}$$