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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 \(\vec{v} \times \vec{d}\):
   \[
   \vec{v} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & -10 & 2 \\ -5 & -3 & 0 \end{vmatrix}
   = \hat{i} ( (-10)(0) - (-3)(2) ) - \hat{j} ( (10)(0) - (-5)(2) ) + \hat{k} ( (10)(-3) - (-10)(-5) )
   \]
   \[
   = \hat{i}(6) - \hat{j}(-10) + \hat{k}(-30 - 50)
   = \langle 6, 10, -80 \rangle.
   \]

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.

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

Learn how to calculate the shortest distance from the point (3, -7, -3) to the line with parametric equation \(\vec{a}(t) = \langle -5, -3, 0 \rangle t + \langle -7, 3, -5 \rangle\). This problem involves vector algebra, including the use of cross products and the distance formula in 3D geometry. Ideal for undergraduate students.

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