To find the equation of the line that passes through the point \((0, 1, -1)\), lies in the plane given by \(4x - 2y + 7z + 9 = 0\), and is parallel to the plane \(2x - 3y + 9z = 0\), we can follow these steps:
### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \(Ax + By + Cz + D = 0\) is \((A, B, C)\).
For the plane \(P: 2x - 3y + 9z = 0\), the normal vector \(n_1\) is:
\[
n_1 = (2, -3, 9)
\]
For the plane \(4x - 2y + 7z + 9 = 0\), the normal vector \(n_2\) is:
\[
n_2 = (4, -2, 7)
\]
### Step 2: Find the direction ratios of the line
Since the line is parallel to the plane \(P\), its direction ratios must be perpendicular to the normal vector of that plane. Let the direction ratios of the line be \((a, b, c)\). The condition for perpendicularity is given by the dot product:
\[
n_1 \cdot (a, b, c) = 0
\]
This gives us the equation:
\[
2a - 3b + 9c = 0 \quad \text{(1)}
\]
### Step 3: Use the condition for the line to lie in the second plane
The line must also lie in the plane \(4x - 2y + 7z + 9 = 0\). Therefore, the direction ratios must satisfy:
\[
n_2 \cdot (a, b, c) = 0
\]
This gives us the equation:
\[
4a - 2b + 7c = 0 \quad \text{(2)}
\]
### Step 4: Solve the system of equations
Now we have two equations:
1. \(2a - 3b + 9c = 0\)
2. \(4a - 2b + 7c = 0\)
We can solve these equations simultaneously.
From equation (1):
\[
2a = 3b - 9c \quad \Rightarrow \quad a = \frac{3b - 9c}{2} \quad \text{(3)}
\]
Substituting (3) into equation (2):
\[
4\left(\frac{3b - 9c}{2}\right) - 2b + 7c = 0
\]
This simplifies to:
\[
6b - 18c - 2b + 7c = 0
\]
\[
4b - 11c = 0 \quad \Rightarrow \quad b = \frac{11}{4}c \quad \text{(4)}
\]
Substituting (4) back into (3):
\[
a = \frac{3\left(\frac{11}{4}c\right) - 9c}{2}
\]
\[
= \frac{\frac{33}{4}c - 9c}{2} = \frac{\frac{33}{4}c - \frac{36}{4}c}{2} = \frac{-3c/4}{2} = -\frac{3c}{8} \quad \text{(5)}
\]
### Step 5: Express the direction ratios
We can express \(a\), \(b\), and \(c\) in terms of \(c\):
\[
a = -\frac{3}{8}c, \quad b = \frac{11}{4}c, \quad c = c
\]
### Step 6: Write the equation of the line
Using the point \((0, 1, -1)\) and the direction ratios \((-3, 22, 8)\) (taking \(c = 8\) for simplicity), we can write the parametric equations of the line:
\[
\frac{x - 0}{-3} = \frac{y - 1}{22} = \frac{z + 1}{8}
\]
Thus, the equation of the line is:
\[
\frac{x}{-3} = \frac{y - 1}{22} = \frac{z + 1}{8}
\]
