To find the distance between a point and a line in 3D, we can use the following formula:

$d = \frac{| \mathbf{b} \times (\mathbf{a_0} - \mathbf{r_0}) |}{| \mathbf{b} |}$

Where:

• $\mathbf{r_0}$ is the given point, $(-3, 6, 3)$.
• $\mathbf{a_0}$ is a point on the line, which can be taken as $(6, -6, -5)$ from the equation of the line.
• $\mathbf{b}$ is the direction vector of the line, which is $(1, -3, -2)$.
• $d$ is the distance from the point to the line.

Steps:

1. Find the vector $\mathbf{a_0} - \mathbf{r_0}$.
2. Compute the cross product $\mathbf{b} \times (\mathbf{a_0} - \mathbf{r_0})$.
3. Find the magnitude of the cross product and divide it by the magnitude of $\mathbf{b}$.

Let's solve it step by step.

1. Vector $\mathbf{a_0} - \mathbf{r_0}$: $\mathbf{a_0} - \mathbf{r_0} = (6, -6, -5) - (-3, 6, 3) = (6 - (-3), -6 - 6, -5 - 3) = (9, -12, -8)$

2. Cross product $\mathbf{b} \times (\mathbf{a_0} - \mathbf{r_0})$: $\mathbf{b} = (1, -3, -2)$ Using the formula for the cross product:

   \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & -2 \\ 9 & -12 & -8 \end{vmatrix}$$ Now, expand the determinant to get the cross product.

Let me calculate this.The cross product $\mathbf{b} \times (\mathbf{a_0} - \mathbf{r_0})$ is $(0, -10, 15)$.

The magnitude of the cross product is approximately $18.03$, and the magnitude of the direction vector $\mathbf{b}$ is approximately $3.74$.

Thus, the distance from the point $(-3, 6, 3)$ to the line is:

$d \approx \frac{18.03}{3.74} \approx 4.82$

The distance is approximately $4.82$ units.

Do you need further details or have any questions?

Here are some related questions:

1. How do you find the projection of a point onto a line?
2. What is the equation for the shortest distance between two parallel lines?
3. How do you compute the angle between two vectors?
4. What is the formula for the distance between a point and a plane?
5. How does the cross product relate to the area of a parallelogram?

Tip: The cross product of two vectors results in a vector perpendicular to both, making it useful in determining areas and distances in 3D geometry.