The distance between the line $r=2i-2j+3k+\lambda \left(i-j+4k\right)$ and the plane $r·\left(i+5j+k\right)=5$

The given line is $r=2i-2j+3k+\lambda \left(i-j+4k\right)$ and the given plane is $r·\left(i+5j+k\right)=5$

since $\left(i-j+4k\right)·\left(i+5j+k\right)=0$

The given line and the plane are parallel

The distance between the line and the plane is equal to the perpendicular distance from any point on the line to the plane 

Hence, the required distance is  equal to the perpendicualr distance from the point $\left(2,-2,3\right)$ to the plane $x+5y+z=5$

Therefore, $\begin{array}{rcl}d& =& \left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|\\ & =& \left|\frac{2-10+3-5}{\sqrt{1+25+1}}\right|\\ & =& \boxed{\left|\frac{10}{3\sqrt{3}}\right|}\end{array}$