The equation of the plane perpendicular to the line `(x-1)/1=(y-2)/(-1)=(z+1)/2` and passing through the point (2,3,1), is

Options

A. `vecr.(hati+hatj+2hatk)=1`
B. `vecr.(hati-hatj+2hatk)=1`
C. `vecr.(hati-hatj+2hatk)=7`
D. `vecr.(hati+hatj-2hatk)=10`

Explanation

To find the equation of the plane that is perpendicular to the given line and passes through the point (2, 3, 1), we will follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in the symmetric form:
\[ \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z+1}{2} \]
From this, we can extract the direction ratios of the line, which are \(1, -1, 2\).

### Step 2: Determine the normal vector of the plane
Since the plane is perpendicular to the line, the direction ratios of the line will be the same as the normal vector of the plane. Therefore, the normal vector \( \mathbf{n} \) of the plane is:
\[ \mathbf{n} = (1, -1, 2) \]

### Step 3: Use the point-normal form of the plane equation
The point-normal form of the equation of a plane is given by:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \((n_1, n_2, n_3)\) are the components of the normal vector and \((x_0, y_0, z_0)\) is a point on the plane.

Substituting the normal vector \((1, -1, 2)\) and the point \((2, 3, 1)\):
\[ 1(x - 2) - 1(y - 3) + 2(z - 1) = 0 \]

### Step 4: Simplify the equation
Expanding the equation:
\[ (x - 2) - (y - 3) + 2(z - 1) = 0 \]
\[ x - 2 - y + 3 + 2z - 2 = 0 \]
Combining like terms:
\[ x - y + 2z - 1 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ x - y + 2z - 1 = 0 \]