Equation of the passing through the origin and perpendicular to the planes `x+2y+z=1`, `3x-4y+z=5` is

Options

A. `x+2y-5z=0`
B. `x-2y-3z=0`
C. `x-2y+5z=0`
D. `3x+y-5z=0`

Explanation

To find the equation of a plane that passes through the origin and is perpendicular to the given planes \(x + 2y + z = 1\) and \(3x - 4y + z = 5\), we can follow these steps:

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

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

For the second plane \(3x - 4y + z = 5\), the normal vector \(\mathbf{n_2}\) is:
\[ \mathbf{n_2} = \langle 3, -4, 1 \rangle \]

### Step 2: Find the cross product of the normal vectors
To find a direction vector that is perpendicular to both planes, we compute the cross product \(\mathbf{n_1} \times \mathbf{n_2}\).

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

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

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 2 & 1 \\ -4 & 1 \end{vmatrix} = (2)(1) - (1)(-4) = 2 + 4 = 6\)
2. \(\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = (1)(1) - (1)(3) = 1 - 3 = -2\)
3. \(\begin{vmatrix} 1 & 2 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (2)(3) = -4 - 6 = -10\)

Thus, we have:
\[ \mathbf{n_1} \times \mathbf{n_2} = 6\mathbf{i} + 2\mathbf{j} - 10\mathbf{k} = \langle 6, 2, -10 \rangle \]

### Step 3: Write the equation of the plane
The equation of a plane can be written in the form:
\[ Ax + By + Cz = D \]
Since the plane passes through the origin, \(D = 0\). Therefore, the equation becomes:
\[ 6x + 2y - 10z = 0 \]

### Step 4: Simplify the equation
We can simplify this equation by dividing all terms by 2:
\[ 3x + y - 5z = 0 \]

### Conclusion
The required equation of the plane that passes through the origin and is perpendicular to the given planes is:
\[ 3x + y - 5z = 0 \]