If the line, `(x-3)/2=(y+2)/(-1)=(z+4)/3` lies in the place, `l x+m y-z=9` , then `l^2+m^2` is equal to: (1) 26 (2) 18 (3) 5 (4) 2

Options

A. `26`
B. `18`
C. `5`
D. `2`

Answer

To solve the problem, we need to analyze the given line and the plane, and then find the values of \( l \) and \( m \) such that the line lies in the plane. Finally, we will compute \( l^2 + m^2 \).

### Step 1: Identify the direction ratios of the line
The line is given by the equation:
\[ \frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3} \]
From this, we can identify the direction ratios of the line as \( (2, -1, 3) \).

### Step 2: Identify a point on the line
To find a point on the line, we can set the parameter \( t = 0 \):
\[ x = 3 + 2(0) = 3, \quad y = -2 + (-1)(0) = -2, \quad z = -4 + 3(0) = -4 \]
Thus, a point \( P \) on the line is \( (3, -2, -4) \).

### Step 3: Substitute the point into the plane equation
The equation of the plane is given by:
\[ l x + m y - z = 9 \]
Substituting the coordinates of point \( P(3, -2, -4) \) into the plane equation:
\[ l(3) + m(-2) - (-4) = 9 \]
This simplifies to:
\[ 3l - 2m + 4 = 9 \]
Subtracting 4 from both sides gives:
\[ 3l - 2m = 5 \quad \text{(Equation 1)} \]

### Step 4: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be represented as \( (l, m, -1) \). Since the line is parallel to the vector \( (2, -1, 3) \), the normal vector must be perpendicular to this vector.

### Step 5: Set up the dot product equation
The dot product of the direction vector of the line and the normal vector must equal zero:
\[ (2, -1, 3) \cdot (l, m, -1) = 0 \]
Calculating the dot product gives:
\[ 2l - m - 3 = 0 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( 3l - 2m = 5 \)
2. \( 2l - m - 3 = 0 \)

From Equation 2, we can express \( m \) in terms of \( l \):
\[ m = 2l - 3 \]
Substituting this into Equation 1:
\[ 3l - 2(2l - 3) = 5 \]
This simplifies to:
\[ 3l - 4l + 6 = 5 \]
Combining like terms gives:
\[ -l + 6 = 5 \]
Thus:
\[ -l = -1 \implies l = 1 \]

### Step 7: Find \( m \)
Substituting \( l = 1 \) back into the expression for \( m \):
\[ m = 2(1) - 3 = 2 - 3 = -1 \]

### Step 8: Calculate \( l^2 + m^2 \)
Now we can compute \( l^2 + m^2 \):
\[ l^2 + m^2 = 1^2 + (-1)^2 = 1 + 1 = 2 \]

### Conclusion
Thus, the value of \( l^2 + m^2 \) is:
\[ \boxed{2} \]