Equations of the line passing through `(1,1,1)` and perpendicular to the plane `2x+3y+z+5=0` are

Options

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

Answer

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

### Step 1: Identify the normal vector of the plane
The coefficients of \( x \), \( y \), and \( z \) in the plane equation \( 2x + 3y + z + 5 = 0 \) represent the components of the normal vector to the plane. Thus, the normal vector \( \vec{n} \) is given by:
\[ \vec{n} = (2, 3, 1) \]

### Step 2: Use the point and the direction ratios to form the equation of the line
Since the line is perpendicular to the plane, its direction ratios will be the same as those of the normal vector. Therefore, the direction ratios \( (l, m, n) \) of the line are:
\[ l = 2, \quad m = 3, \quad n = 1 \]

### Step 3: Write the equation of the line
The equation of a line in three-dimensional space that passes through a point \( (x_1, y_1, z_1) \) and has direction ratios \( (l, m, n) \) can be expressed as:
\[ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \]
Substituting the point \( (1, 1, 1) \) and the direction ratios \( (2, 3, 1) \):
\[ \frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1} \]

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