Find the equation of the line parallel to $2\overline{i}-\overline{j}+2\overline{k}$  and which passes through point $\text{'}A\text{'}$  $(3\overline{i}+\overline{j}-\overline{k})$. If $P$ is a point on the line such that $AP=15$. Find the position vectors of $P$.

let $\overline{a}=3\overline{i}+\overline{j}-\overline{k}, \overline{b}=2\overline{i}-\overline{j}+2\overline{k}$

The vector equation of the line passing through the point whose position vector is $\overline{a}$ and parallel to the vector $\overline{b}$ is $\overline{r}=\overline{a}+t\overline{b}$

$\Rightarrow \overline{r}=(3\overline{i}+\overline{j}-\overline{k})+t(2\overline{i}-\overline{j}+2\overline{k})$

where $t\in R$

Let $\overline{OP}=x\overline{i}+y\overline{j}+z\overline{k}$  be a point on the line

Now $\overline{AP}=\overline{OP}-\overline{OA}$

$(x-3)i+(y-1)\overline{j}+(z+1)k$

Also $\overline{AP}$ and $\overline{b}$ are parallel

$\Rightarrow \overline{AP}=\lambda \overline{b}\Rightarrow \left|\overline{AP}\right|=\lambda \left|\overline{b}\right|.$

$15=\left|\lambda \right|3\Rightarrow \lambda =\pm 5$

$\therefore$$\overline{OP}=\overline{OA}+\overline{AP}=(3,1,-1)\pm 5(2,-1,2)$

$=(13,-4,9)$ or $(-7,6,-1)$