What is the closest point to a plane? I can't solve this question: Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Find the point in this plane that is closest to $(0,3,-1)$.
I don't know how to do this question. Any hints/solutions? If it is vectors, then I am pretty sure I don't know how to do it. :(

It is the orthogonal projection $H$  of the $M(0,3,-1)$ onto the plane defined by the points $A(-3,4,-2)$, $B(1,4,0)$ and $C(3,2,-1)$.  


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*If the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are orthogonal, it is given by the formula
$$\overrightarrow{AH}=\frac{\overrightarrow{AM}\cdot\overrightarrow{AB}}{\overrightarrow{AB}\cdot\overrightarrow{AB}}\,\overrightarrow{AB}+\frac{\overrightarrow{AM}\cdot\overrightarrow{AC}}{\overrightarrow{AC}\cdot\overrightarrow{AC}}\,\overrightarrow{AC}.\tag{1}$$
To end determining the point $H$, just calculate 
$$\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{AH}.$$

*If the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are not orthogonal, you first have to deduce an orthogonal basis of the plane $\bigl\langle\overrightarrow{AB}, \overrightarrow{AC}\bigr\rangle$ by Gram-Schmidt process: if $$\overrightarrow{AC'}=\overrightarrow{AC}-\frac{\overrightarrow{AM}\cdot\overrightarrow{AB}}{\overrightarrow{AB}\cdot\overrightarrow{AB}}\,\overrightarrow{AB},$$
the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC'}$ are orthogonal, and you can apply  formula  $(1)$.
