To solve the problem, we need to find the line of intersection of the two given planes and analyze the statements provided.
### Step 1: Write down the equations of the planes.
The equations of the planes are:
1. \( P_1: 3x - 6y - 2z = 15 \)
2. \( P_2: 2x + y - 2z = 5 \)
### Step 2: Find the normal vectors of the planes.
The normal vector of plane \( P_1 \) is \( \vec{n_1} = \langle 3, -6, -2 \rangle \) and for plane \( P_2 \), the normal vector is \( \vec{n_2} = \langle 2, 1, -2 \rangle \).
### Step 3: Find the direction vector of the line of intersection.
The direction vector \( \vec{d} \) of the line of intersection of the two planes can be found using the cross product of the normal vectors:
\[
\vec{d} = \vec{n_1} \times \vec{n_2}
\]
Calculating the cross product:
\[
\vec{d} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & -6 & -2 \\
2 & 1 & -2
\end{vmatrix}
= \hat{i}((-6)(-2) - (-2)(1)) - \hat{j}((3)(-2) - (-2)(2)) + \hat{k}((3)(1) - (-6)(2))
\]
\[
= \hat{i}(12 + 2) - \hat{j}(-6 + 4) + \hat{k}(3 + 12)
\]
\[
= \hat{i}(14) - \hat{j}(-2) + \hat{k}(15)
\]
\[
= 14\hat{i} + 2\hat{j} + 15\hat{k}
\]
### Step 4: Find a point on the line of intersection.
To find a point on the line, we can set \( z = 0 \) and solve the equations:
1. From \( P_1: 3x - 6y = 15 \)
2. From \( P_2: 2x + y = 5 \)
Substituting \( z = 0 \) into the equations:
1. \( 3x - 6y = 15 \) simplifies to \( x - 2y = 5 \)
2. \( 2x + y = 5 \)
Now we can solve these two equations. From \( x - 2y = 5 \), we can express \( x \) in terms of \( y \):
\[
x = 2y + 5
\]
Substituting into the second equation:
\[
2(2y + 5) + y = 5 \implies 4y + 10 + y = 5 \implies 5y + 10 = 5 \implies 5y = -5 \implies y = -1
\]
Substituting \( y = -1 \) back to find \( x \):
\[
x = 2(-1) + 5 = 3
\]
Thus, when \( z = 0 \), we have the point \( (3, -1, 0) \).
### Step 5: Write the parametric equations of the line.
Using the point \( (3, -1, 0) \) and the direction vector \( \langle 14, 2, 15 \rangle \), the parametric equations of the line are:
\[
x = 3 + 14t, \quad y = -1 + 2t, \quad z = 0 + 15t
\]
### Step 6: Analyze the statements.
- **Statement 1**: The parametric equations of the line intersection of the given planes are \( x = 3 + 14t, y = 2t, z = 15t \). This statement is **false** because the correct equation for \( y \) should be \( y = -1 + 2t \).
- **Statement 2**: The vector \( 14\hat{i} + 2\hat{j} + 15\hat{k} \) is parallel to the line of intersection of the given planes. This statement is **true** since it is the direction vector we found.
### Conclusion:
The correct option is that Statement 1 is false and Statement 2 is true.
