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

$\frac{10}{3\sqrt{3}}$

Clearly, the given line passes through the point $\vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and is parallel to the vector $\vec{b} = \hat{i} - \hat{j} + 4\hat{k}.$

The plane is normal to the vector $\vec{n} = \hat{i}+5\hat{j} + \hat{k}.$

We have, $\vec{b} .\vec{n} = 1- 5 + 4 = 0.$

So, the line is parallel to the plane.

∴ Required distance

= Length of the perpendicular from a point on the line to the given plane.

= Length of the perpendicular from $(2\hat{i} - 2\hat{j} = 3\hat{k})$ to the given plane.

$= \begin{vmatrix} \frac{(2\hat{i}-2\hat{j}+3\hat{k}).(\hat{i}+5\hat{j}+\hat{k})-5}{\sqrt{1+25+1}}\end{vmatrix}= \begin{vmatrix} \frac{2-10+3-5}{3\sqrt{3}}\end{vmatrix} =\frac{10}{3\sqrt{3}}$