If the projection of point $P(\vec{p})$ on the plane $\vec{r} . \vec{n}=q$ is the point $S(\vec{S})$, then:
Answer & explanation
Correct answer: option 2
We have
$\vec{s}-\vec{p}=\lambda \vec{n}$ and $\vec{s} . \vec{n}=q$
$\Rightarrow(\lambda \vec{n}+\vec{p}) . \vec{n}=q$
$\Rightarrow \vec{s}=\vec{p}+\frac{(q-\vec{p} . \vec{n}) \vec{n}}{|\vec{n}|^2}$
Hence (2) is correct answer.