Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{ı}}-\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}}+\mathrm{\lambda }(3\widehat{\mathrm{ı}}+4\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}})$ and the plane $\overrightarrow{\mathrm{r}}\cdot (\widehat{\mathrm{ı}}-\widehat{\mathrm{ȷ}}+\widehat{\mathrm{k}})=5$.

Given: $\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{ı}}-\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}}+\mathrm{\lambda }(3\widehat{\mathrm{ı}}+4\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}})\dots \dots \text{ (i) }$
Coordinates of any point on this line are 
$(2+3\mathrm{\lambda })\widehat{\mathrm{i}}+(-1+4\mathrm{\lambda })\widehat{\mathrm{j}}+(2+2\mathrm{\lambda })\widehat{\mathrm{k}}$
Equation of plane is 
$\overrightarrow{\mathrm{r}}\cdot (\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}})=5 .....\left(\mathrm{ii}\right)$
Since, the point on line lies on the plane, 
$\overrightarrow{\mathrm{r}}\cdot (\widehat{\mathrm{i}}-\widehat{\mathrm{i}}+\widehat{\mathrm{k}})=5$
$\begin{array}{l}\left[\right(2+3\mathrm{\lambda })\widehat{\mathrm{i}}+(2+4\mathrm{\lambda })\widehat{\mathrm{j}}+(2+2\mathrm{\lambda }\left)\right](\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}})=5\\ \Rightarrow (2+3\mathrm{\lambda })-(-1+4\mathrm{\lambda })+(2+2\mathrm{\lambda })=5\\ \Rightarrow \mathrm{\lambda }+5=5\\ \Rightarrow \mathrm{\lambda }=0\\ \end{array}$
So equation of line is
$\begin{array}{l}\overrightarrow{\mathrm{r}}=(2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})+0(3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})\\ \Rightarrow \overrightarrow{\mathrm{r}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+2\widehat{\mathrm{k}} ....\left(\mathrm{iii}\right)\end{array}$
Let, point of intersection be $(\mathrm{x}, \mathrm{y}, \mathrm{z})$
$\therefore \overrightarrow{\mathrm{r}}=\mathrm{x}\widehat{\mathrm{i}}+\mathrm{y}\widehat{\mathrm{j}}+\mathrm{z}\widehat{\mathrm{k} .....\left(\mathrm{iv}\right)}$
From equations (iii) and (iv)
$\mathrm{x}=2,\mathrm{y}=-1,\mathrm{z}=2$
$\therefore$ Point of intersection is (2,-1,2).
Distance between points (2,-1,2) and (-1,-5,-10) is
$\begin{array}{l}=\sqrt{(-1-2{)}^2+(-5+1{)}^2+(-10-2{)}^2}\\ =\sqrt{(-3{)}^2+(-4{)}^2+(-12{)}^2}\\ =\sqrt{9+16+144}=\sqrt{169}=13\text{ units }\end{array}$
Therefore, the distance of the point (– 1, – 5, – 10) from the point of intersection of the line 
$\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{ı}}-\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}}+\mathrm{\lambda }(3\widehat{\mathrm{ı}}+4\widehat{\mathrm{ȷ}}+2\widehat{\mathrm{k}})$ and the plane $\overrightarrow{\mathrm{r}}\cdot (\widehat{\mathrm{ı}}-\widehat{\mathrm{ȷ}}+\widehat{\mathrm{k}})=5$ is 13 units.