Vector equation of the given line is,

$r=\left(3\hat{i}+\hat{j}-\hat{k}\right)+t\left(2\hat{i}-\hat{j}+2\hat{k}\right)$

$\Rightarrow \overrightarrow{BA}=t\left(2\hat{i}-\hat{j}+2\hat{k}\right)=2\hat{i}t-\hat{j}t+2\hat{k}t$

$\Rightarrow 18=\sqrt{4t^2+t^2+4t^2}=\pm 3t$

$\therefore t=\pm 6$

$\therefore \overrightarrow{OA}=\left(3\hat{i}+\hat{j}-\hat{k}\right)\pm 6\left(2\hat{i}-\hat{j}+2\hat{k}\right)$

$=3\hat{i}+12\hat{i}+\hat{j}-6\hat{j}-\hat{k}+12\hat{k}$

$=15\hat{i}-5\hat{j}+11\hat{k}$

or $=3\hat{i}-12\hat{i}+\hat{j}+6\hat{j}-\hat{k}-12\hat{k}$

$=-9\hat{i}+7\hat{j}-13\hat{k}$