To solve the problem, we need to find the equation of the plane \( p \) that passes through the point \( (3, 7, -7) \) and contains the line given by the parametric equations \( \frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1} \). Finally, we will calculate the square of the distance \( d^2 \) from the origin to the plane.
### Step 1: Identify a point on the line
The line can be expressed in parametric form as:
\[
x = 2 - 3t, \quad y = 3 + 2t, \quad z = -2 + t
\]
For \( t = 0 \), we can find a point on the line:
\[
(2, 3, -2)
\]
### Step 2: Determine the direction vector of the line
From the line's equation, we can identify the direction vector:
\[
\vec{d} = (-3, 2, 1)
\]
### Step 3: Find the direction vector between the two points
Now, we have two points: \( A(3, 7, -7) \) and \( B(2, 3, -2) \). We can find the direction vector \( \vec{AB} \):
\[
\vec{AB} = (2 - 3, 3 - 7, -2 + 7) = (-1, -4, 5)
\]
### Step 4: Find the normal vector of the plane
The normal vector \( \vec{n} \) of the plane can be found using the cross product of the two direction vectors \( \vec{d} \) and \( \vec{AB} \):
\[
\vec{n} = \vec{d} \times \vec{AB} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-3 & 2 & 1 \\
-1 & -4 & 5
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i} \left( 2 \cdot 5 - 1 \cdot (-4) \right) - \hat{j} \left( -3 \cdot 5 - 1 \cdot (-1) \right) + \hat{k} \left( -3 \cdot (-4) - 2 \cdot (-1) \right)
\]
\[
= \hat{i} (10 + 4) - \hat{j} (-15 + 1) + \hat{k} (12 + 2)
\]
\[
= \hat{i} (14) + \hat{j} (14) + \hat{k} (14)
\]
Thus, the normal vector is:
\[
\vec{n} = (14, 14, 14)
\]
### Step 5: Write the equation of the plane
Using the point-normal form of the plane equation:
\[
14(x - 3) + 14(y - 7) + 14(z + 7) = 0
\]
This simplifies to:
\[
x + y + z - 3 = 0
\]
### Step 6: Calculate the distance from the origin to the plane
The distance \( d \) from the origin \( (0, 0, 0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Here, \( A = 1, B = 1, C = 1, D = -3 \):
\[
d = \frac{|1 \cdot 0 + 1 \cdot 0 + 1 \cdot 0 - 3|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}
\]
### Step 7: Find \( d^2 \)
Finally, we calculate \( d^2 \):
\[
d^2 = (\sqrt{3})^2 = 3
\]
### Final Answer
Thus, the value of \( d^2 \) is:
\[
\boxed{3}
\]
