To solve this problem, we need to find the coordinates of the point $p$ on line $L$ that is closest to the point $(-16, 19, -17)$.

Step 1: Find the Direction Vector of the Line

The line $L$ passes through points $(12, 20, -14)$ and $(-9, 13, 11)$. The direction vector $\vec{d}$ of line $L$ is found by subtracting these points: $\vec{d} = (-9 - 12, 13 - 20, 11 - (-14)) = (-21, -7, 25)$

Step 2: Parameterize the Line $L$

A point on the line $L$ can be represented as: $(x, y, z) = (12, 20, -14) + t(-21, -7, 25)$ where $t$ is a parameter. Thus, the coordinates of any point $(x, y, z)$ on $L$ are: $x = 12 - 21t$ $y = 20 - 7t$ $z = -14 + 25t$

Step 3: Find the Closest Point on $L$ to $(-16, 19, -17)$

The point $p = (p_1, p_2, p_3)$ on $L$ that is closest to $(-16, 19, -17)$ will be the orthogonal projection of $(-16, 19, -17)$ onto $L$.

To find $t$ for the closest point, we need to set up the vector from $(12, 20, -14)$ to $(-16, 19, -17)$, which is: $\vec{v} = (-16 - 12, 19 - 20, -17 + 14) = (-28, -1, -3)$

Now, project $\vec{v}$ onto $\vec{d}$ using the formula: $t = \frac{\vec{v} \cdot \vec{d}}{\vec{d} \cdot \vec{d}}$

1. Calculate $\vec{v} \cdot \vec{d}$: $\vec{v} \cdot \vec{d} = (-28)(-21) + (-1)(-7) + (-3)(25) = 588 + 7 - 75 = 520$

2. Calculate $\vec{d} \cdot \vec{d}$: $\vec{d} \cdot \vec{d} = (-21)^2 + (-7)^2 + (25)^2 = 441 + 49 + 625 = 1115$

3. Substitute to find $t$: $t = \frac{520}{1115} \approx 0.466$

Step 4: Substitute $t$ Back to Find $p$

Now we can substitute $t \approx 0.466$ into the parametric equations for $(x, y, z)$: $p_1 = 12 - 21(0.466) \approx 2.206$ $p_2 = 20 - 7(0.466) \approx 16.738$ $p_3 = -14 + 25(0.466) \approx -2.85$

Final Answer

The coordinates of the point $p$ on $L$ closest to $(-16, 19, -17)$ are approximately: $p = (2.206, 16.738, -2.85)$

Would you like more detailed steps on any part of this solution, or do you have any specific questions?

Here are five additional questions related to this topic:

1. How would the solution change if we had different points on the line?
2. How can we generalize this method for any line and external point in 3D space?
3. What is the significance of the direction vector in determining the line’s path?
4. How is the concept of projection used to find the closest point?
5. What would be the closest point if the external point were on the line?

Tip: Understanding vector projection can be very useful for solving problems involving distances from points to lines or planes.