How do you detect if a point is in a plane? Let's say we have 3 points: (-2,7,4), (-4,5,2), (3,8,5) and we want to see if a fourth point, (2,6,3), is in the plane that the previous 3 points made. How would I go about doing this?
 A: Let $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ be the three given points, and $(a,b,c)$ be the fourth.  A little linear algebra shows that $(a,b,c)$ is in the plane if and only if the following matrix has rank 2 (assuming of course that the three given points are not collinear):
$$
\left[\begin{array}{ccc}
  x_2-x_1 & y_2-y_1 & z_2-z_1 \\
  x_3-x_1 & y_3-y_1 & z_3-z_1 \\
  a-x_1 & b-y_1 & c-z_1 \end{array}\right]
$$
Hope this helps!
A: Let $x_1,x_2,x_3 \in \mathbb{R}^3$ be three non-collinear points. Let $n = (x_2-x_1) \times (x_3-x_1)$, $\hat{n} = \frac{1}{\|n\|} n$, and $d = \langle x_1, \hat{n} \rangle$. Then $|\langle x, \hat{n} \rangle -d|$ is the distance from the point $x$ to the plane spanned by $x_1,x_2,x_3 $. Typically, you need to decide numerically whether or not the point is actually on the plane.
In your case we can perform the calculations exactly: $n = (0, -8, 8)^T$, $\hat{n} = \frac{1}{8\sqrt{2}} (0, -8, 8)^T$,
$d = -\frac{3}{\sqrt{2}}$.
If we let $x = (2,6,3)^T$, we have $\langle x, \hat{n} \rangle = -\frac{3}{\sqrt{2}}$, hence the distance to the plane = $|-\frac{3}{\sqrt{2}}+\frac{3}{\sqrt{2}}| = 0$, so the point is on the plane.
In fact, with a little more calculation, you can show that $x = \frac{1}{8} (-11 x_1 + 9 x_2 + 10 x_3)$, and since the coefficients sum to $1$, $x$ lies in what is known as the affine hull of $x_1,x_2,x_3$, which in this case is the plane containing the points.
To complete the connection, you can show that $\langle x-x_1, n \rangle = \det A$, where $A = \begin{bmatrix} x_2-x_1 & x_3-x_1 & x - x_1 \end{bmatrix}$. Note that $A$ is the transpose of the matrix in Shaun's answer. The catch here is that you need to scale by $\frac{1}{\|n\|}$ to get a numerically meaningful answer.
A: Use the given three points to find the equation of the plane, $ax+by+cz=d$. Then plug in the fourth point and see if it satisfies the equation.
