Find the equation of the plane which passes through the point `(3, 4, -5)` and contains the lines `(x+1)/(2)=(y-1)/(3)=(z+2)/(-1)`

Answer

To find the equation of the plane that passes through the point \( (3, 4, -5) \) and contains the line given by the equation \( \frac{x+1}{2} = \frac{y-1}{3} = \frac{z+2}{-1} \), we can follow these steps:

### Step 1: Identify a point on the line
The line can be expressed in parametric form. Let \( t \) be the parameter. Then we can write:
- \( x = 2t - 1 \)
- \( y = 3t + 1 \)
- \( z = -t - 2 \)

To find a specific point on the line, we can choose \( t = 0 \):
- \( x = 2(0) - 1 = -1 \)
- \( y = 3(0) + 1 = 1 \)
- \( z = -0 - 2 = -2 \)

Thus, the point \( (-1, 1, -2) \) lies on the line.

### Step 2: Find the direction ratios of the line
From the parametric equations, we can see that the direction ratios of the line are \( (2, 3, -1) \).

### Step 3: Find the normal vector of the plane
The plane contains the line, so its normal vector must be perpendicular to the direction ratios of the line. Let the normal vector be \( (a, b, c) \). The condition for perpendicularity is given by:
\[ 2a + 3b - c = 0 \quad \text{(1)} \]

### Step 4: Use the point-normal form of the plane
The equation of a plane in point-normal form is given by:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Substituting the point \( (3, 4, -5) \):
\[ a(x - 3) + b(y - 4) + c(z + 5) = 0 \quad \text{(2)} \]

### Step 5: Substitute the point on the line into the plane equation
Substituting the point \( (-1, 1, -2) \) into equation (2):
\[ a(-1 - 3) + b(1 - 4) + c(-2 + 5) = 0 \]
This simplifies to:
\[ -4a - 3b + 3c = 0 \quad \text{(3)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( 2a + 3b - c = 0 \) (from the direction ratios)
2. \( -4a - 3b + 3c = 0 \) (from the point on the line)

From equation (1), we can express \( c \):
\[ c = 2a + 3b \quad \text{(4)} \]
Substituting (4) into equation (3):
\[ -4a - 3b + 3(2a + 3b) = 0 \]
This simplifies to:
\[ -4a - 3b + 6a + 9b = 0 \]
\[ 2a + 6b = 0 \implies a + 3b = 0 \implies a = -3b \]

### Step 7: Express \( a, b, c \) in terms of \( b \)
Let \( b = k \) (a parameter), then:
\[ a = -3k, \quad b = k, \quad c = 2(-3k) + 3k = -6k + 3k = -3k \]

### Step 8: Substitute back into the plane equation
Substituting \( a, b, c \) into equation (2):
\[ -3k(x - 3) + k(y - 4) - 3k(z + 5) = 0 \]
Dividing through by \( k \) (assuming \( k \neq 0 \)):
\[ -3(x - 3) + (y - 4) - 3(z + 5) = 0 \]
Expanding this gives:
\[ -3x + 9 + y - 4 - 3z - 15 = 0 \]
Combining like terms:
\[ -3x + y - 3z - 10 = 0 \]
Rearranging gives:
\[ 3x - y + 3z + 10 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ 3x - y + 3z + 10 = 0 \]