Distance of $P(\vec{p})$ from the plane $\vec{r} . \vec{n}=0$ is:

$\frac{|\vec{p} . \vec{n}|}{|\vec{n}|}$

Let $Q(\vec{q})$ be the foot of altitude drawn from P to the plane $\vec{r} . \vec{n}=0$,

$\Rightarrow \vec{q}-\vec{p}=\lambda \vec{n}$

$\Rightarrow \vec{q}=\vec{p}+\lambda \vec{n}$

Also $\vec{q} . \vec{n}=0$

$\Rightarrow(\vec{p}+\lambda \vec{n}) . \vec{n}=0$

$\Rightarrow \lambda=-\frac{(\vec{p} . \vec{n})}{|\vec{n}|^2}$

$\Rightarrow \vec{q}-\vec{p}=-\frac{(\vec{p} . \vec{n})}{|\vec{n}|^2} \vec{n}$

∴ Required distance

$=|\vec{q}-\vec{p}|$

$=\frac{|\vec{p} . \vec{n}|}{|\vec{n}|}=|\vec{p} . \hat{n}|$

Hence (3) is correct answer.