The problem asks to find the distance from the point $(3, -7, -3)$ to the line $\vec{a}(t) = \langle -5, -3, 0 \rangle t + \langle -7, 3, -5 \rangle$.

Steps to solve:

1. Point on the Line: The parametric equation of the line is given by: $\vec{a}(t) = \langle -5, -3, 0 \rangle t + \langle -7, 3, -5 \rangle = \langle -5t - 7, -3t + 3, -5 \rangle.$ Here, the point on the line depends on the parameter $t$.

2. Direction Vector of the Line: The direction vector of the line is $\langle -5, -3, 0 \rangle$.

3. Vector from Point to a Point on the Line: The point $(3, -7, -3)$ is not on the line. To find the vector from this point to a specific point on the line (say when $t = 0$, $\vec{a}(0) = \langle -7, 3, -5 \rangle$), we can calculate the vector: $\vec{v} = \langle 3, -7, -3 \rangle - \langle -7, 3, -5 \rangle = \langle 3 + 7, -7 - 3, -3 + 5 \rangle = \langle 10, -10, 2 \rangle.$

4. Projection of Vector onto the Line: The formula for the distance between a point and a line in 3D is given by: $d = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|},$ where $\vec{v} = \langle 10, -10, 2 \rangle$ is the vector from the point to a point on the line, and $\vec{d} = \langle -5, -3, 0 \rangle$ is the direction vector of the line.

5. Cross Product: Let's compute the cross product $\vec{v} \times \vec{d}$:

   = \hat{i} ( (-10)(0) - (-3)(2) ) - \hat{j} ( (10)(0) - (-5)(2) ) + \hat{k} ( (10)(-3) - (-10)(-5) )$$ $$= \hat{i}(6) - \hat{j}(-10) + \hat{k}(-30 - 50) = \langle 6, 10, -80 \rangle.$$

6. Magnitude of the Cross Product: Now, calculate the magnitude of $\vec{v} \times \vec{d}$: $|\vec{v} \times \vec{d}| = \sqrt{6^2 + 10^2 + (-80)^2} = \sqrt{36 + 100 + 6400} = \sqrt{6536}.$

7. Magnitude of the Direction Vector: The magnitude of the direction vector $\vec{d} = \langle -5, -3, 0 \rangle$ is: $|\vec{d}| = \sqrt{(-5)^2 + (-3)^2 + 0^2} = \sqrt{25 + 9} = \sqrt{34}.$

8. Final Distance: Now, the distance $d$ is: $d = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{6536}}{\sqrt{34}} = \sqrt{\frac{6536}{34}} = \sqrt{192} \approx 13.86.$

The distance from the point $(3, -7, -3)$ to the line is approximately 13.86 units.

Would you like a deeper breakdown or explanation of any steps?

Here are 5 related questions to deepen your understanding:

1. How do you find the equation of a line in 3D space?
2. What is the geometric interpretation of the cross product?
3. How do you project a vector onto another vector in 3D?
4. What is the formula for the distance between two parallel lines?
5. How can you compute the shortest distance from a point to a plane?

Tip: For vector calculations like cross products, visualizing the vectors helps in understanding the spatial relationships better.