To solve the problem, we need to show that the line of intersection of the two given planes is coplanar with the given line and then find the equation of the plane that contains both lines.
### Step 1: Identify the equations of the planes
The equations of the planes are:
1. \( P_1: x + 2y + 3z = 8 \)
2. \( P_2: 2x + 4y + 4z = 11 \)
### Step 2: Find the line of intersection of the two planes
To find the line of intersection of the two planes, we can express one variable in terms of the others. We can solve the system of equations formed by the two planes.
From \( P_1 \):
\[
x + 2y + 3z = 8 \quad \text{(1)}
\]
From \( P_2 \):
\[
2x + 4y + 4z = 11 \quad \text{(2)}
\]
We can simplify equation (2) by dividing everything by 2:
\[
x + 2y + 2z = \frac{11}{2} \quad \text{(3)}
\]
Now we have:
1. \( x + 2y + 3z = 8 \)
2. \( x + 2y + 2z = \frac{11}{2} \)
Now, subtract equation (3) from equation (1):
\[
(x + 2y + 3z) - (x + 2y + 2z) = 8 - \frac{11}{2}
\]
This simplifies to:
\[
z = 8 - \frac{11}{2} = \frac{16}{2} - \frac{11}{2} = \frac{5}{2}
\]
Substituting \( z = \frac{5}{2} \) back into equation (1):
\[
x + 2y + 3\left(\frac{5}{2}\right) = 8
\]
\[
x + 2y + \frac{15}{2} = 8
\]
\[
x + 2y = 8 - \frac{15}{2} = \frac{16}{2} - \frac{15}{2} = \frac{1}{2}
\]
So, we can express \( x \) in terms of \( y \):
\[
x = \frac{1}{2} - 2y
\]
Thus, we have:
\[
x = \frac{1}{2} - 2y, \quad z = \frac{5}{2}
\]
### Step 3: Parametric equations of the line of intersection
Let \( y = t \) (a parameter). Then:
\[
x = \frac{1}{2} - 2t, \quad y = t, \quad z = \frac{5}{2}
\]
The parametric equations of the line of intersection are:
\[
\begin{align*}
x &= \frac{1}{2} - 2t \\
y &= t \\
z &= \frac{5}{2}
\end{align*}
\]
### Step 4: Equation of the given line
The given line is:
\[
\frac{x + 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{3}
\]
We can express this in parametric form by letting \( k \) be the parameter:
\[
\begin{align*}
x &= k - 1 \\
y &= 2k - 1 \\
z &= 3k - 1
\end{align*}
\]
### Step 5: Check if the lines are coplanar
To check if the two lines are coplanar, we can use the scalar triple product. We need to find the direction vectors of both lines and a vector connecting a point on one line to a point on the other line.
The direction vector of the line of intersection (from Step 3) is:
\[
\mathbf{d_1} = \langle -2, 1, 0 \rangle
\]
The direction vector of the given line (from Step 4) is:
\[
\mathbf{d_2} = \langle 1, 2, 3 \rangle
\]
Choose a point on the line of intersection (let \( t = 0 \)):
\[
P_1 = \left(\frac{1}{2}, 0, \frac{5}{2}\right)
\]
Choose a point on the given line (let \( k = 0 \)):
\[
P_2 = (-1, -1, -1)
\]
Now, find the vector \( \mathbf{P_1P_2} \):
\[
\mathbf{P_1P_2} = \left(-1 - \frac{1}{2}, -1 - 0, -1 - \frac{5}{2}\right) = \left(-\frac{3}{2}, -1, -\frac{7}{2}\right)
\]
### Step 6: Compute the scalar triple product
The scalar triple product is given by:
\[
\mathbf{d_1} \cdot (\mathbf{d_2} \times \mathbf{P_1P_2})
\]
Calculating \( \mathbf{d_2} \times \mathbf{P_1P_2} \):
\[
\mathbf{d_2} = \langle 1, 2, 3 \rangle, \quad \mathbf{P_1P_2} = \left(-\frac{3}{2}, -1, -\frac{7}{2}\right)
\]
Using the determinant:
\[
\mathbf{d_2} \times \mathbf{P_1P_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-\frac{3}{2} & -1 & -\frac{7}{2}
\end{vmatrix}
\]
Calculating this determinant gives:
\[
\mathbf{d_2} \times \mathbf{P_1P_2} = \mathbf{i}\left(2 \cdot -\frac{7}{2} - 3 \cdot -1\right) - \mathbf{j}\left(1 \cdot -\frac{7}{2} - 3 \cdot -\frac{3}{2}\right) + \mathbf{k}\left(1 \cdot -1 - 2 \cdot -\frac{3}{2}\right)
\]
\[
= \mathbf{i}(-7 + 3) - \mathbf{j}\left(-\frac{7}{2} + \frac{9}{2}\right) + \mathbf{k}(-1 + 3)
\]
\[
= \mathbf{i}(-4) - \mathbf{j}\left(\frac{2}{2}\right) + \mathbf{k}(2) = \langle -4, -1, 2 \rangle
\]
Now compute \( \mathbf{d_1} \cdot \langle -4, -1, 2 \rangle \):
\[
\mathbf{d_1} = \langle -2, 1, 0 \rangle
\]
\[
\mathbf{d_1} \cdot \langle -4, -1, 2 \rangle = (-2)(-4) + (1)(-1) + (0)(2) = 8 - 1 + 0 = 7
\]
Since the scalar triple product is not zero, the lines are coplanar.
### Step 7: Find the equation of the plane containing both lines
The equation of the plane can be expressed in the form:
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
Using the normal vector \( \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} \):
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & 1 & 0 \\
1 & 2 & 3
\end{vmatrix}
\]
Calculating this determinant gives:
\[
\mathbf{n} = \mathbf{i}(1 \cdot 3 - 0 \cdot 2) - \mathbf{j}(-2 \cdot 3 - 0 \cdot 1) + \mathbf{k}(-2 \cdot 2 - 1 \cdot 1)
\]
\[
= \mathbf{i}(3) - \mathbf{j}(-6) + \mathbf{k}(-4 - 1)
\]
\[
= \langle 3, 6, -5 \rangle
\]
Using point \( P_1 = \left(\frac{1}{2}, 0, \frac{5}{2}\right) \):
\[
3\left(x - \frac{1}{2}\right) + 6(y - 0) - 5\left(z - \frac{5}{2}\right) = 0
\]
Expanding this gives:
\[
3x - \frac{3}{2} + 6y - 5z + \frac{25}{2} = 0
\]
\[
3x + 6y - 5z + \frac{22}{2} = 0
\]
\[
3x + 6y - 5z + 11 = 0
\]
### Final Solution
The equation of the plane containing both lines is:
\[
3x + 6y - 5z + 11 = 0
\]
