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To solve the problem of finding the scalar parametric equations for the line formed by the intersection of two planes, we need to:

1. **Identify the given planes**: 
   - The first plane is \( 3x + 3y + 2z + 5 = 0 \).
   - The second plane is \( 4x - 3y + 2z - 1 = 0 \).

2. **Find the direction vector of the line**: 
   - The direction vector of the line formed by the intersection of two planes is given by the cross product of the normal vectors of the planes.
   - The normal vector of the first plane is \( \mathbf{n_1} = \langle 3, 3, 2 \rangle \).
   - The normal vector of the second plane is \( \mathbf{n_2} = \langle 4, -3, 2 \rangle \).

   Let's calculate the cross product \( \mathbf{n_1} \times \mathbf{n_2} \).

3. **Find a point on the line**:
   - Substitute one of the variables (typically \( z \)) as a parameter (let \( z = t \)), and solve the system of equations to find the corresponding \( x \) and \( y \) values.

4. **Write the parametric equations**:
   - Once the direction vector and a point on the line are known, the parametric equations can be written in the form:
     \[
     \begin{aligned}
     x(t) &= x_0 + at \\
     y(t) &= y_0 + bt \\
     z(t) &= z_0 + ct
     \end{aligned}
     \]
     where \( \langle x_0, y_0, z_0 \rangle \) is a point on the line and \( \langle a, b, c \rangle \) is the direction vector.

I'll calculate the necessary steps.The direction vector of the line formed by the intersection of the two planes is \( \langle 12, 2, -21 \rangle \).

Next, we need to find a specific point on the line. We can do this by setting one of the variables (say \( z = t \)) to a specific value and solving the system of equations.

I'll proceed with that calculation.The solutions for \( x \) and \( y \) in terms of the parameter \( z = t \) are:

\[
x(t) = -\frac{4}{7}t - \frac{4}{7}
\]
\[
y(t) = -\frac{2}{21}t - \frac{23}{21}
\]

Now we can write the parametric equations for the line:

\[
\begin{aligned}
x(t) &= -\frac{4}{7}t - \frac{4}{7} \\
y(t) &= -\frac{2}{21}t - \frac{23}{21} \\
z(t) &= t
\end{aligned}
\]

These are the parametric equations of the line formed by the intersection of the two planes.

Would you like any more details, or do you have any questions? Here are five related questions to explore further:

1. How would the parametric equations change if we chose a different parameter for \( z \)?
2. Can you verify the solution by substituting these equations back into the original plane equations?
3. What is the geometric significance of the direction vector found?
4. How would you find the point of intersection between this line and another plane?
5. What happens if the two planes are parallel or identical?

**Tip:** When solving parametric equations, always check if the chosen parameterization leads to simplified and meaningful results.

Find a set of scalar parametric equations for the line formed by the two intersecting planes. The planes are given as:

3x + 3y + 2z + 5 = 0
4x - 3y + 2z - 1 = 0

Learn how to find the scalar parametric equations for a line formed by the intersection of two planes. This detailed explanation involves vector cross products, parametric equations, and finding a direction vector, suitable for high school and college-level math students.

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