To find the distance between a point and a line given by parametric equations, we'll use the following steps:

1. Express the line in vector form: The parametric equations for the line are: $x = 2t, \quad y = t - 3, \quad z = 2t + 2$ This can be written as: $\mathbf{r}(t) = (2t, t - 3, 2t + 2)$ The direction vector $\mathbf{v}$ of the line is given by the coefficients of $t$: $\mathbf{v} = \langle 2, 1, 2 \rangle$ A point on the line when $t = 0$ is: $\mathbf{r}_0 = (0, -3, 2)$

2. Find the vector from the point to a point on the line: Let the given point be $\mathbf{P} = (1, -1, 5)$. The vector from $\mathbf{r}_0$ to $\mathbf{P}$ is: $\mathbf{P} - \mathbf{r}_0 = \langle 1 - 0, -1 - (-3), 5 - 2 \rangle = \langle 1, 2, 3 \rangle$

3. Compute the cross product of this vector with the direction vector: The cross product $\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)$ is: $\mathbf{v} \times \langle 1, 2, 3 \rangle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}$ $= \mathbf{i} \left(1 \cdot 3 - 2 \cdot 2\right) - \mathbf{j} \left(2 \cdot 3 - 2 \cdot 1\right) + \mathbf{k} \left(2 \cdot 2 - 1 \cdot 1\right)$ $= \mathbf{i} (3 - 4) - \mathbf{j} (6 - 2) + \mathbf{k} (4 - 1)$ $= \langle -1, -4, 3 \rangle$

4. Calculate the magnitude of the cross product: $|\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)| = \sqrt{(-1)^2 + (-4)^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$

5. Compute the magnitude of the direction vector: $|\mathbf{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

6. Calculate the distance: The distance $d$ from the point to the line is given by: $d = \frac{|\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)|}{|\mathbf{v}|} = \frac{\sqrt{26}}{3}$ $d \approx \frac{5.099}{3} \approx 1.700$

So, the distance between the point $(1, -1, 5)$ and the line is approximately $1.700$ units.

Would you like more details or have any questions?

Here are some related questions:

1. How is the cross product of two vectors calculated?
2. What is the significance of the magnitude of a vector?
3. How can you derive the distance formula from a point to a line?
4. What are the parametric equations of a line in three-dimensional space?
5. How does the direction vector of a line relate to its parametric equations?
6. Can the distance between a point and a line ever be negative?
7. What role does the dot product play in geometry?
8. How do you find the shortest distance between two skew lines?

Tip: Always verify your vector calculations step by step to avoid errors, especially in cross products.