Point On a Plane Closest to a Point 
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*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 $(0,3,-1)$.





*Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Let $\ell$ be the line containing the points $(2, 4, -3)$ and $(-1, -1, -9)$.
Find the intersection of the plane $\mathcal{P}$ and the line $\ell$.



I don't know where to start.  I don't know any formulas for calculating a closest point.

Problem 1.
The equation of the plane containing the points $(−3,4,−2)$, $(1,4,0)$ and $(3,2,−1)$ is given by 
\begin{align*}
\begin{vmatrix}
x & y & z & 1 \\ 
-3 & 4 & -2 & 1 \\
1 & 4 & 0 & 1 \\
3 & 2 & -1 & 1
\end{vmatrix}
&= 0
\end{align*}
or, $x + 2y - 2z - 9 = 0$
The vector normal to this plane is given by $(1, 2, -2)$
The point in this plane that is closest to $(0,3,−1)$ must be the foot of the perpendicular from this point to the above plane.
The equation of the perpendicular is
$\frac{x - 0}{1} = \frac{y-3}{2} = \frac{z+1}{-2} = k$
or, $x = k, y = 2k + 3, z = -2k-1$
Putting this in the equation of the plane,
$k + 2(2k+3) + 2(2k+1) - 9 = 0$
or, $k = \frac{1}{9}$
Hence the point is $(\frac{1}{9}, \frac{29}{9}, -\frac{11}{9})$
Problem 2.
The equation of the line containing the points $(2,4,−3)$ and $(−1,−1,−9)$ is given by 
$\frac{x - 2}{-1-2} = \frac{y-4}{-1-4} = \frac{z+3}{-9+3} = t$
or, $x = -3t+2, y = -5t+4, z = -6t-3$
Putting this in the equation of the plane,
$-3t+2 + 2(-5t+4) + 2(6t+3) - 9 = 0$
or, $t = 7$
So the point of intersection is $(-19, -31, -45)$
[Please check the calculations]
