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

Options

A. `(x-1)/3=(y+1)/3=(z-0)/(-1)`
B. `(x-1)/3=(y+1)/1=(z-0)/(-1)`
C. `(x-1)/3=(y+1)/1=(z-0)/1`
D. `(x-1)/3=(y+1)/2=(z-0)/1`

Explanation

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

### Step 1: Identify the point and direction ratios
The line we want to find passes through the point \( P(1, -1, 0) \). The direction ratios of the given line can be extracted from the equation.

### Step 2: Convert the given line equation to standard form
The given line equation is:
\[ \frac{x-2}{3} = \frac{2y+1}{2} = \frac{5-z}{-1} \]
This can be rewritten in the standard form:
\[ \frac{x-2}{3} = \frac{y + \frac{1}{2}}{1} = \frac{z - 5}{-1} \]
From this, we can identify the direction ratios of the line as \( (3, 1, -1) \).

### Step 3: Write the direction vector
The direction vector \( \mathbf{b} \) of the line is:
\[ \mathbf{b} = 3\mathbf{i} + 1\mathbf{j} - 1\mathbf{k} \]

### Step 4: Use the point-direction form of the line
The equation of a line in space can be expressed in the form:
\[ \frac{x - x_1}{a_1} = \frac{y - y_1}{a_2} = \frac{z - z_1}{a_3} \]
where \( (x_1, y_1, z_1) \) is a point on the line and \( (a_1, a_2, a_3) \) are the direction ratios.

### Step 5: Substitute the values
Here, \( (x_1, y_1, z_1) = (1, -1, 0) \) and the direction ratios \( (a_1, a_2, a_3) = (3, 1, -1) \). Substituting these values into the equation gives:
\[ \frac{x - 1}{3} = \frac{y + 1}{1} = \frac{z - 0}{-1} \]

### Final Equation
Thus, the equation of the required line is:
\[ \frac{x - 1}{3} = y + 1 = -z \]