I think I can clear up some misunderstanding. A line contains more than just two points. A line is made up of infinitely many points. It is however true that a line is determined by 2 points, namely just extend the line segment connecting those two points.
Similarly a plane is determined by 3 non-co-linear points. In this case the three points are a point from each line and the point of intersection. We are not creating a new point when the lines intersect, the point was already there.
This is not the same thing as saying that there are 5 points because there are two from each line and the point from their intersection.
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]
Best Answer
Find an equation of the plane, and one of the line.
$V_1 := (4,0,2)$ and $V_2 := (-2,2,1)$ are two linearly independent vectors in $P$, so $N := V_1 \times V_2 = (-4,-8,8)$ is a normal vector to $P$. Let $n = (1,2,-2)$ (just for simplicity). Then $n$ is also a normal to $P$, and $P$'s equation is given by:
$$n \cdot (x - 1, y-4, z) = 0$$
Therefore:
$$ P: x + 2y -2z = 9$$
$U:= (3,5,6)$ is a direction vector of $\ell$, and $\ell$ passes through $A:= (2,4,-3)$. We know:
$$\ell : \begin{cases} x = x_A + x_U t \\ y = y_A + y_U t \\ z = z_A + z_U t \end{cases}$$
Thus:
$$\ell : \begin{cases} x= 3t + 2 \\ y = 5t + 4 \\ z = 6t - 3 \end{cases}$$
Let $M=(a,b,c)$ be the intersection point between $P$ and $\ell$. $M$ satisfies the equations of each: $P$ and $\ell$, so that for some $t$: $a= 3t + 2$, $b= 5t + 4$ and $c = 6t -3$, and $a+2b -2c = 9$. Hence, $3t + 2 + 10t + 8 -12t + 6 = 9$, so $t = -7$.
Therefore, $M = (-19, -31, -45)$.