Equation of a line passing through `(-1,2,-3)` and perpendicular to the plane `2x+3y+z+5=0` is

Options

A. `(x-1)/(-1)=(y+2)/1=(z-3)/(-1)`
B. `(x+1)/(-1)=(y-2)/1=(z+3)/1`
C. `(x+1)/2=(y-2)/3=(z+3)/1`
D. none of these

Explanation

To find the equation of a line passing through the point \((-1, 2, -3)\) and perpendicular to the plane given by the equation \(2x + 3y + z + 5 = 0\), we can follow these steps:

### Step 1: Identify the point through which the line passes
The point given is \((-1, 2, -3)\). This will be our point \((x_1, y_1, z_1)\).

### Step 2: Determine the normal vector of the plane
The equation of the plane is given as \(2x + 3y + z + 5 = 0\). The coefficients of \(x\), \(y\), and \(z\) in this equation represent the components of the normal vector to the plane. Therefore, the normal vector \(\mathbf{n}\) is:
\[ \mathbf{n} = (2, 3, 1) \]
This normal vector will also serve as the direction vector \(\mathbf{b}\) of the line since the line is perpendicular to the plane.

### Step 3: Write the parametric equations of the line
The parametric equations of a line can be expressed in the form:
\[ \frac{x - x_1}{b_1} = \frac{y - y_1}{b_2} = \frac{z - z_1}{b_3} = \lambda \]
Substituting the values:
- \(x_1 = -1\), \(y_1 = 2\), \(z_1 = -3\)
- \(b_1 = 2\), \(b_2 = 3\), \(b_3 = 1\)

We get:
\[ \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z + 3}{1} = \lambda \]

### Step 4: Write the final equation in symmetric form
From the above parametric equations, we can express the symmetric form of the line as:
\[ \frac{x + 1}{2} = \frac{y - 2}{3} = z + 3 \]

### Final Answer
Thus, the equation of the line passing through the point \((-1, 2, -3)\) and perpendicular to the plane \(2x + 3y + z + 5 = 0\) is:
\[ \frac{x + 1}{2} = \frac{y - 2}{3} = z + 3 \]

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