A line is fixed by a point and a direction along it. A plane is fixed by a point and a direction perpendicular to it, called a normal vector. Once you know the normal, the dot product turns "lies in the plane" into an equation, and the cross product produces a normal from two directions in the plane.
🔑 Key idea
The plane through $P_0 = (x_0, y_0, z_0)$ with normal $\mathbf{n} = \langle a, b, c \rangle$:
$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0, \quad\text{or}\quad ax + by + cz = d.$$
The coefficients of $x$, $y$, $z$ are a normal vector. Through three points $A$, $B$, $C$: $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$.
💡 Why it works
Pick any point $X = (x, y, z)$. It lies in the plane exactly when the vector $\overrightarrow{P_0X}$ runs along the plane, which means it is perpendicular to $\mathbf{n}$: $\mathbf{n} \cdot \overrightarrow{P_0X} = 0$. Writing that dot product out in components is the equation above. For three points, $\overrightarrow{AB}$ and $\overrightarrow{AC}$ both lie in the plane, and their cross product is perpendicular to both, so it is perpendicular to the whole plane.
The plane $x + 2y + 2z = 4$. The red normal $\mathbf{n} = \langle 1, 2, 2 \rangle$ sticks straight out; the gold vectors lie in the plane, and each has dot product ${}0$ with $\mathbf{n}$. Turn the picture until the plane is seen edge-on: the normal is then perpendicular to that edge.
Worked examples
✏️ Worked example: Point and normal
Find the plane through $(2, -1, 3)$ with normal $\langle 4, 1, -2 \rangle$.
${}4(x - 2) + 1(y + 1) - 2(z - 3) = 0$.
Expand: ${}4x - 8 + y + 1 - 2z + 6 = 0$, so ${}4x + y - 2z = 1$.
Check the point: ${}8 - 1 - 6 = 1$.
Answer: ${}4x + y - 2z = 1$
✏️ Worked example: Three points
Find the plane through $A(1, 0, 0)$, $B(0, 2, 0)$, $C(0, 0, 3)$.
Two planes are parallel when their normals are parallel, and perpendicular when their normals are perpendicular. The angle between two planes is the angle between their normals (take the acute one).
⚠️ Common mistake
Using a vector that lies in the plane as the normal. For the plane through $A$, $B$, $C$, the vector $\overrightarrow{AB}$ runs along the plane; putting its components in front of $x$, $y$, $z$ gives a different plane, one that crosses this plane. The normal comes from a cross product (or from the coefficients of a given equation). Check: your normal dotted with $\overrightarrow{AB}$ and with $\overrightarrow{AC}$ must be ${}0$, and all three points must satisfy your equation.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.