Given line passing through $A$ having position vector $\overrightarrow{a}=\hat{i}+2\hat{j}-2\hat{k}$

And line parallel to $\overrightarrow{b}=2\hat{i}-3\hat{j}-6\hat{k}$

Therefore, equation of line is 

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}-2\hat{k}\right)+\lambda \left(2\hat{i}-3\hat{j}-6\hat{k}\right)$

$\Rightarrow \overrightarrow{r}=\left(1-2\lambda \right)\hat{i}+\left(2-3\lambda \right)\hat{j}+\left(-2-6\lambda \right)\hat{k}$

Let, the position vector of $P$ is $\left(1-2\lambda \right)\hat{i}+\left(2-3\lambda \right)\hat{j}+\left(-2-6\lambda \right)\hat{k}$.

$\because \left|\overrightarrow{AP}\right|=21$

$\Rightarrow \left|-2\lambda \hat{i}-3\lambda \hat{j}-6\lambda \hat{k}\right|=21$

$\Rightarrow \sqrt{4{\lambda }^2+9{\lambda }^2+36{\lambda }^2}=21$

$\Rightarrow 7\left|\lambda \right|=21$

$\therefore \lambda =\pm 3$

When $\lambda =3$, then

$P\equiv -5\hat{i}-7\hat{j}-20\hat{k}$

When $\lambda =-3$, then

$P\equiv 7\hat{i}+11\hat{j}+16\hat{k}$