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}. (\hat{i}-5\hat{j}+\hat{k}) - 4 = 0 $ is :

$\frac{5\sqrt{3}}{3}$

The correct answer is Option (4) → $\frac{5\sqrt{3}}{3}$

$(\hat i+\hat j+4\hat k).(\hat i-5\hat j+\hat k)=0$

line || plane

for $λ=0$ points on line (2, 2, -3)

distance from plane is $\frac{(2×1-5×2-3×1-4)}{\sqrt{1+25+1}}$

$=\frac{5\sqrt{3}}{3}$ units