To find the orthogonal component of vector $\mathbf{a}$ with respect to vector $\mathbf{b}$, we first need to determine the projection of $\mathbf{a}$ onto $\mathbf{b}$.

1. Calculate the dot product $\mathbf{a} \cdot \mathbf{b}$: $\mathbf{a} \cdot \mathbf{b} = (-7)(2) + (-7)(0) + (5)(-3) = -14 + 0 - 15 = -29$

2. Calculate the dot product of $\mathbf{b}$ with itself: $\mathbf{b} \cdot \mathbf{b} = (2)(2) + (0)(0) + (-3)(-3) = 4 + 0 + 9 = 13$

3. Find the projection of $\mathbf{a}$ onto $\mathbf{b}$: $\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} = \frac{-29}{13} \mathbf{b}$ $= \frac{-29}{13} (2, 0, -3) = \left(\frac{-58}{13}, 0, \frac{87}{13}\right)$

4. Calculate the orthogonal component $\mathbf{a}_{\perp}$: $\mathbf{a}_{\perp} = \mathbf{a} - \text{proj}_{\mathbf{b}} \mathbf{a}$ $= (-7, -7, 5) - \left(\frac{-58}{13}, 0, \frac{87}{13}\right) = \left(-7 + \frac{58}{13}, -7 - 0, 5 - \frac{87}{13}\right)$

   Now, convert $-7$ to a fraction: $-7 = \frac{-91}{13}$ So, $\mathbf{a}_{\perp} = \left(\frac{-91 + 58}{13}, -7, 5 - \frac{87}{13}\right) = \left(\frac{-33}{13}, -7, \frac{65 - 87}{13}\right) = \left(\frac{-33}{13}, -7, \frac{-22}{13}\right)$

Thus, the orthogonal component of $\mathbf{a}$ with respect to $\mathbf{b}$ is: $\left(\frac{-33}{13}, -7, \frac{-22}{13}\right)$

Would you like further details or have any questions about this process? Here are some related questions you might consider:

1. How do you visualize orthogonal components in a geometric context?
2. What is the significance of orthogonal vectors in physics?
3. Can you explain how to find the orthogonal component in two dimensions?
4. What happens if two vectors are parallel in terms of their orthogonal components?
5. How can the concept of orthogonal projection be applied in machine learning?

Tip: When calculating projections and orthogonal components, always double-check your vector operations to ensure accuracy!