Equation of the plane passing through the point (1,1,1) and perpendicular to each of the planes` x+2y+3z=7` and `2x-3y+4z=0`l is

Options

A. `17x-2y+7z=12`
B. `17x+2y-7z=12`
C. `17x+2y+7z=12`
D. `17x-2y-7z=12`

Explanation

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

### Step 1: Identify the normal vectors of the given planes
The equations of the given planes are:
1. \( x + 2y + 3z = 7 \)
2. \( 2x - 3y + 4z = 0 \)

From these equations, we can extract the normal vectors:
- For the first plane, the normal vector \( \mathbf{n_1} = (1, 2, 3) \).
- For the second plane, the normal vector \( \mathbf{n_2} = (2, -3, 4) \).

### Step 2: Calculate the cross product of the normal vectors
To find the normal vector of the required plane, we need to compute the cross product of \( \mathbf{n_1} \) and \( \mathbf{n_2} \):

\[ \mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
- For \( \mathbf{i} \): \( (2 \cdot 4) - (3 \cdot -3) = 8 + 9 = 17 \)
- For \( \mathbf{j} \): \( (1 \cdot 4) - (3 \cdot 2) = 4 - 6 = -2 \) (note the negative sign in front)
- For \( \mathbf{k} \): \( (1 \cdot -3) - (2 \cdot 2) = -3 - 4 = -7 \)

Thus, we have:
\[ \mathbf{n} = (17, 2, -7) \]

### Step 3: Use the point-normal form of the plane equation
The equation of a plane can be expressed in the form:
\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \]
where \( \mathbf{r} = (x, y, z) \) is a general point on the plane, \( \mathbf{r_0} = (1, 1, 1) \) is a point on the plane, and \( \mathbf{n} = (17, 2, -7) \) is the normal vector.

Substituting these values into the equation:
\[ (17, 2, -7) \cdot ((x, y, z) - (1, 1, 1)) = 0 \]

This expands to:
\[ 17(x - 1) + 2(y - 1) - 7(z - 1) = 0 \]

### Step 4: Simplify the equation
Expanding this gives:
\[ 17x - 17 + 2y - 2 - 7z + 7 = 0 \]
Combining like terms:
\[ 17x + 2y - 7z - 12 = 0 \]

Rearranging gives us the final equation of the plane:
\[ 17x + 2y - 7z = 12 \]

### Final Answer
The equation of the required plane is:
\[ 17x + 2y - 7z = 12 \]