The equation of the line passing through `(-4, 3, 1),` parallel to the plane `x +2y-z-5=0` and intersecting the line `(x+1)/-3=(y-3)/2=(z-2)/-1`

A.`(x+4)/1=(y-3)/1=(z-1)/3`
B. `(x+4)/3=(y-3)/(-1)=(z-1)/1`
C. `(x-4)/2=(y+3)/1=(z+1)/4`
D. `(x+4)/(-1)=(y-3)/1=(z-1)/1`

Options

A. `(x+4)/1=(y-3)/1=(z-1)/3`
B. `(x+4)/3=(y-3)/(-1)=(z-1)/1`
C. `(x-4)/2=(y+3)/1=(z+1)/4`
D. `(x+4)/(-1)=(y-3)/1=(z-1)/1`

Explanation

To solve the problem, we need to find the equation of a line that passes through the point \((-4, 3, 1)\), is parallel to the plane given by the equation \(x + 2y - z - 5 = 0\), and intersects the line defined by the equation \(\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}\).

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as \(x + 2y - z - 5 = 0\). The normal vector \(\mathbf{n}\) to the plane can be derived from the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n} = (1, 2, -1) \]

### Step 2: Find the direction ratios of the line
Since the line we are looking for is parallel to the plane, its direction ratios must be perpendicular to the normal vector of the plane. Let the direction ratios of the line be \((a, b, c)\). The condition for being perpendicular is given by the dot product:
\[ 1a + 2b - 1c = 0 \]
This simplifies to:
\[ a + 2b - c = 0 \tag{1} \]

### Step 3: Find the direction ratios of the given line
The given line is represented as:
\[ \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1} \]
From this, we can extract the direction ratios of the line:
\[ (-3, 2, -1) \]

### Step 4: Set up the intersection condition
For the line we are looking for to intersect the given line, we can set up the determinant condition. The points on the two lines must satisfy the following determinant:
\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ -3 & 2 & -1 \\ a & b & c \end{vmatrix} = 0 \]
Where \((x_1, y_1, z_1) = (-4, 3, 1)\) and \((x_2, y_2, z_2)\) are points on the second line.

### Step 5: Substitute the points
Substituting the points into the determinant gives:
\[ \begin{vmatrix} -1 + 4 & 3 - 3 & 2 - 1 \\ -3 & 2 & -1 \\ a & b & c \end{vmatrix} = 0 \]
This simplifies to:
\[ \begin{vmatrix} 3 & 0 & 1 \\ -3 & 2 & -1 \\ a & b & c \end{vmatrix} = 0 \]

### Step 6: Calculate the determinant
Calculating the determinant:
\[ 3 \begin{vmatrix} 2 & -1 \\ b & c \end{vmatrix} - 0 + 1 \begin{vmatrix} -3 & 2 \\ a & b \end{vmatrix} = 0 \]
This expands to:
\[ 3(2c + b) + (-3b - 2a) = 0 \]
Rearranging gives:
\[ 6c + 3b - 3b - 2a = 0 \implies 6c - 2a = 0 \implies a = 3c \tag{2} \]

### Step 7: Substitute into equation (1)
Substituting \(a = 3c\) into equation (1):
\[ 3c + 2b - c = 0 \implies 2b + 2c = 0 \implies b = -c \]

### Step 8: Find the ratios
Now we have:
\[ a = 3c, \quad b = -c, \quad c = c \]
This gives us the ratios:
\[ a : b : c = 3c : -c : c \implies 3 : -1 : 1 \]

### Step 9: Write the equation of the line
Using the point \((-4, 3, 1)\) and the direction ratios \((3, -1, 1)\), we can write the equation of the line as:
\[ \frac{x + 4}{3} = \frac{y - 3}{-1} = \frac{z - 1}{1} \]

### Conclusion
Thus, the equation of the line is:
\[ \frac{x + 4}{3} = \frac{y - 3}{-1} = \frac{z - 1}{1} \]
The correct option is **B**.