• # question_answer The distance between the line$\overrightarrow{a}=2\hat{i}-2\hat{j}+3\hat{k}$ $\lambda (\hat{i}+5\hat{j}+\hat{k})=5$and the plane$\overrightarrow{r}.(\hat{i}+5\hat{j}+\hat{k})=5$is A)  $\frac{10}{9}$ B)  $\frac{10}{3\sqrt{3}}$ C)  $\frac{3}{10}$ D)  $\frac{10}{3}$

Here, $(\hat{i}-\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0$ So, the equation of the line parallel to the plane is The  general  point  on this  line  is$(\lambda +2,-\lambda -2,4\lambda +3)$for$\lambda =0$the point on the line is  and distance from $\overrightarrow{r}.(\hat{i}+5\hat{j}+\hat{k})=5$or$x+5y+z=5$ $\therefore$ $d=\left| \frac{2+5(-2)+3-5}{\sqrt{1+25+1}} \right|$ $\Rightarrow$ $d=\left| \frac{-10}{3\sqrt{3}} \right|=\frac{10}{3\sqrt{3}}$