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

Options

A. 3x – 10y – 2z + 11 = 0
B. 6x – 5y – 2z – 2 = 0
C. 11x + y + 17z + 38 = 0
D. 6x – 5y + 2z + 10 = 0

Answer

To find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the given planes, 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 \(3x + y - 2z = 5\), the normal vector \(\mathbf{n_1}\) is:
\[ \mathbf{n_1} = (3, 1, -2) \]

For the second plane \(2x - 5y - z = 7\), the normal vector \(\mathbf{n_2}\) is:
\[ \mathbf{n_2} = (2, -5, -1) \]

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

Calculating the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 1 & -2 \\ -5 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -2 \\ 2 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 1 \\ 2 & -5 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & -2 \\ -5 & -1 \end{vmatrix} = (1)(-1) - (-2)(-5) = -1 - 10 = -11\)

2. \(\begin{vmatrix} 3 & -2 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (-2)(2) = -3 + 4 = 1\)

3. \(\begin{vmatrix} 3 & 1 \\ 2 & -5 \end{vmatrix} = (3)(-5) - (1)(2) = -15 - 2 = -17\)

Thus, we have:
\[ \mathbf{n} = -11\mathbf{i} - 1\mathbf{j} - 17\mathbf{k} = (-11, -1, -17) \]

### Step 4: Write the equation of the plane
The equation of a plane can be written as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane, and \((n_1, n_2, n_3)\) are the components of the normal vector.

Substituting the point \((1, 2, -3)\) and the normal vector \((-11, -1, -17)\):
\[ -11(x - 1) - 1(y - 2) - 17(z + 3) = 0 \]

### Step 5: Simplifying the equation
Expanding this:
\[ -11x + 11 - y + 2 - 17z - 51 = 0 \]
\[ -11x - y - 17z - 38 = 0 \]
Multiplying through by -1 gives:
\[ 11x + y + 17z + 38 = 0 \]

### Final Answer
The equation of the plane is:
\[ 11x + y + 17z + 38 = 0 \]