The equation of the plane passing through the point (1,1,1) and perpendicular to the planes `2x+y-2z=5 and 3x-6y-2z=7`

Options

A. `14x+2y+15x=31`
B. `14x+2y-15z=1`
C. `14x+2y+15x=3`
D. `14x-2y+15z=27`

Explanation

To find the equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes given by the equations \(2x + y - 2z = 5\) and \(3x - 6y - 2z = 7\), we can follow these steps:

### Step 1: Identify the normal vectors of the given planes
The normal vector of a plane in the form \(Ax + By + Cz = D\) is given by the coefficients \(A\), \(B\), and \(C\).

For the first plane \(2x + y - 2z = 5\), the normal vector \(\mathbf{n_1}\) is:
\[ \mathbf{n_1} = \langle 2, 1, -2 \rangle \]

For the second plane \(3x - 6y - 2z = 7\), the normal vector \(\mathbf{n_2}\) is:
\[ \mathbf{n_2} = \langle 3, -6, -2 \rangle \]

### Step 2: Find the normal vector of the required plane
The normal vector of the required plane, which is perpendicular to both \(\mathbf{n_1}\) and \(\mathbf{n_2}\), can be found using the cross product:
\[ \mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -2 \\ 3 & -6 & -2 \end{vmatrix} \]

Expanding this determinant:
\[ \mathbf{n} = \mathbf{i} \left(1 \cdot (-2) - (-2) \cdot (-6)\right) - \mathbf{j} \left(2 \cdot (-2) - (-2) \cdot 3\right) + \mathbf{k} \left(2 \cdot (-6) - 1 \cdot 3\right) \]
\[ = \mathbf{i} \left(-2 - 12\right) - \mathbf{j} \left(-4 + 6\right) + \mathbf{k} \left(-12 - 3\right) \]
\[ = -14\mathbf{i} - 2\mathbf{j} - 15\mathbf{k} \]

Thus, the normal vector \(\mathbf{n}\) is:
\[ \mathbf{n} = \langle -14, -2, -15 \rangle \]

### Step 3: Use the point-normal form to find the equation of the plane
The equation of a plane in point-normal form is given by:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((n_x, n_y, n_z)\) is the normal vector.

Substituting \((x_0, y_0, z_0) = (1, 1, 1)\) and \((n_x, n_y, n_z) = (-14, -2, -15)\):
\[ -14(x - 1) - 2(y - 1) - 15(z - 1) = 0 \]

### Step 4: Simplify the equation
Expanding this gives:
\[ -14x + 14 - 2y + 2 - 15z + 15 = 0 \]
\[ -14x - 2y - 15z + 31 = 0 \]

Multiplying through by -1 to simplify:
\[ 14x + 2y + 15z = 31 \]

### Final Answer
The equation of the required plane is:
\[ 14x + 2y + 15z = 31 \]