Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line

The equation of the given line is

$\vec{r} \cdot=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$          $\ldots(1)$

The equation of the given plane is

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$                                                                       $\ldots(2)$

Substituting the value of $\vec{r}$ from equation (1) in equation (2), we obtain

$[2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})] \cdot(\hat{i}-\hat{j}+\hat{k})=5$

$\Rightarrow[(3 \lambda+2) \hat{i}+(4 \lambda-1) \hat{j}+(2 \lambda+2) \hat{k}] \cdot(\hat{i}-\hat{j}+\hat{k})=5$

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

$\Rightarrow \lambda=0$

Substituting this value in equation (1), we obtain the equation of the line as

$\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}$

This means that the position vector of the point of intersection of the line and the plane is $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}$

This shows that the point of intersection of the given line and plane is given by the coordinates, $(2,-1,2)$. The point is $(-1,-5,-10)$.

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

$d=\sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}}=\sqrt{9+16+144}=\sqrt{169}=13$