To find the projection of vector $\mathbf{P}$ onto vector $\mathbf{Q}$, you can use the formula:

$\text{Proj}_{\mathbf{Q}} \mathbf{P} = \frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{Q}|^2} \mathbf{Q}$

Here's how we can solve this step by step:

1. Calculate the dot product $\mathbf{P} \cdot \mathbf{Q}$:

$\mathbf{P} \cdot \mathbf{Q} = (1)(-3) + (6)(-5) + (5)(5) = -3 - 30 + 25 = -8$

2. Calculate the magnitude squared of $\mathbf{Q}$:

$|\mathbf{Q}|^2 = (-3)^2 + (-5)^2 + 5^2 = 9 + 25 + 25 = 59$

3. Use the projection formula:

$\text{Proj}_{\mathbf{Q}} \mathbf{P} = \frac{-8}{59} \mathbf{Q} = \frac{-8}{59} \langle -3, -5, 5 \rangle$

4. Multiply the scalar by the vector $\mathbf{Q}$:

$\text{Proj}_{\mathbf{Q}} \mathbf{P} = \left\langle \frac{-8}{59} \times -3, \frac{-8}{59} \times -5, \frac{-8}{59} \times 5 \right\rangle = \left\langle \frac{24}{59}, \frac{40}{59}, \frac{-40}{59} \right\rangle$

So, the projection of vector $\mathbf{P} = \langle 1, 6, 5 \rangle$ onto vector $\mathbf{Q} = \langle -3, -5, 5 \rangle$ is:

$\text{Proj}_{\mathbf{Q}} \mathbf{P} = \left\langle \frac{24}{59}, \frac{40}{59}, \frac{-40}{59} \right\rangle$

Would you like any more details or have any other questions?

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Here are 5 related questions for further exploration:

1. How do you interpret the projection of one vector onto another?
2. What is the difference between the projection and the component of a vector?
3. How do you find the perpendicular component of a vector relative to another vector?
4. How does the projection formula change in two-dimensional space?
5. What are some applications of vector projections in physics or engineering?

Tip: The projection of a vector $\mathbf{P}$ onto another vector $\mathbf{Q}$ shows the shadow or footprint of $\mathbf{P}$ along $\mathbf{Q}$.