To find the equation of a plane containing the lines \( L_1 \) and \( L_2 \), we first need to extract the direction ratios and a point from each line.
### Step 1: Identify the direction ratios and points of the lines
From the equations of the lines:
- For line \( L_1: \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1} \)
- Direction ratios \( a_1 = -3, b_1 = 2, c_1 = 1 \)
- A point on \( L_1 \) can be found by setting the parameter to 0:
- \( x = -1, y = 3, z = -2 \) (Point \( A(-1, 3, -2) \))
- For line \( L_2: \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} \)
- Direction ratios \( a_2 = 1, b_2 = -3, c_2 = 2 \)
- A point on \( L_2 \) can be found by setting the parameter to 0:
- \( x = 0, y = 7, z = -7 \) (Point \( B(0, 7, -7) \))
### Step 2: Calculate the normal vector to the plane
To find the equation of the plane containing both lines, we need a normal vector to the plane. This can be obtained by taking the cross product of the direction ratios of the two lines.
The direction vectors are:
- \( \vec{d_1} = (-3, 2, 1) \)
- \( \vec{d_2} = (1, -3, 2) \)
The cross product \( \vec{n} = \vec{d_1} \times \vec{d_2} \) is calculated as follows:
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-3 & 2 & 1 \\
1 & -3 & 2
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i} \begin{vmatrix} 2 & 1 \\ -3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & 2 \\ 1 & -3 \end{vmatrix}
\]
Calculating each of these determinants:
1. \( \begin{vmatrix} 2 & 1 \\ -3 & 2 \end{vmatrix} = (2)(2) - (1)(-3) = 4 + 3 = 7 \)
2. \( \begin{vmatrix} -3 & 1 \\ 1 & 2 \end{vmatrix} = (-3)(2) - (1)(1) = -6 - 1 = -7 \)
3. \( \begin{vmatrix} -3 & 2 \\ 1 & -3 \end{vmatrix} = (-3)(-3) - (2)(1) = 9 - 2 = 7 \)
Thus,
\[
\vec{n} = 7\hat{i} + 7\hat{j} + 7\hat{k} = (7, 7, 7)
\]
### Step 3: Write the equation of the plane
The equation of a plane can be expressed in the form:
\[
n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0
\]
Using point \( A(-1, 3, -2) \) and normal vector \( (7, 7, 7) \):
\[
7(x + 1) + 7(y - 3) + 7(z + 2) = 0
\]
Simplifying this:
\[
7x + 7 + 7y - 21 + 7z + 14 = 0
\]
\[
7x + 7y + 7z + 0 = 0
\]
Dividing by 7:
\[
x + y + z = 0
\]
### Final Answer
The equation of the plane containing the lines \( L_1 \) and \( L_2 \) is:
\[
x + y + z = 0
\]
