To find the normal vector to two given vectors, you can use the cross product. Let's denote the vectors as follows:

• Vector AB: $\vec{AB} = (-3, -1, 7)$
• Vector AC: $\vec{AC} = (-3, 0, 3)$

The cross product of these two vectors, $\vec{AB} \times \vec{AC}$, will give a vector that is perpendicular to both.

The cross product formula for two vectors $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$ is:

$\vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$

Let's compute it:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -1 & 7 \\ -3 & 0 & 3 \end{vmatrix}$

I'll calculate it step-by-step.

First, expanding the determinant:

$\vec{AB} \times \vec{AC} = \mathbf{i} \begin{vmatrix} -1 & 7 \\ 0 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -3 & 7 \\ -3 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -3 & -1 \\ -3 & 0 \end{vmatrix}$

Let's solve these determinants:

1. For $\mathbf{i}$: $\begin{vmatrix} -1 & 7 \\ 0 & 3 \end{vmatrix} = (-1)(3) - (7)(0) = -3$
2. For $\mathbf{j}$: $\begin{vmatrix} -3 & 7 \\ -3 & 3 \end{vmatrix} = (-3)(3) - (7)(-3) = -9 + 21 = 12$
3. For $\mathbf{k}$: $\begin{vmatrix} -3 & -1 \\ -3 & 0 \end{vmatrix} = (-3)(0) - (-1)(-3) = 0 - 3 = -3$

Thus, the cross product is:

$\vec{AB} \times \vec{AC} = -3\mathbf{i} - 12\mathbf{j} - 3\mathbf{k}$

So the normal vector to the given vectors is:

$\vec{n} = (-3, -12, -3)$

Would you like more details or have any questions?

Here are five related questions:

1. How do you find the magnitude of a normal vector?
2. What is the geometric meaning of a cross product?
3. How do you normalize a vector?
4. Can you find the area of the parallelogram formed by two vectors?
5. What are the properties of a cross product?

Tip: The direction of the cross product depends on the right-hand rule, which determines the orientation of the resulting vector.