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: 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: