18  Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$ and the plane   $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$.

Given,

Equation of a line :

$\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$

Equation of the plane

$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$

Let's first find out the point of intersection of line and plane.

putting the value of $\vec r$ into the equation of a plane from the equation from line

$\left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5$

$(2+3\lambda)-(4\lambda -1)+(2+2\lambda)=5$

$\lambda+5=5$

$\lambda=0$

Now, from the equation, any  point p  in line is

$P=(2+3\lambda,4\lambda-1,2+2\lambda)$

So the point of intersection is

$P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)$

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is

$d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}$

$d=\sqrt{169}=13$

Hence the required distance is 13.

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