Find the equation of a plane passes through the point `(0 ,0,0)` and perpendicular to each to the planes `x+2y-z=1` and `3x-4y+z=5`.

Answer

To find the equation of a plane that passes through the origin (0, 0, 0) 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 represented by the equation \(Ax + By + Cz = D\) is given by the coefficients \(A\), \(B\), and \(C\).

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

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

### Step 2: Find the normal vector of the required plane

The normal vector of the required plane, \(n\), will be perpendicular to both \(n_1\) and \(n_2\). We can find \(n\) by calculating the cross product \(n_1 \times n_2\).

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

Calculating the determinant:

\[ n = \hat{i} \left(2 \cdot 1 - (-1) \cdot (-4)\right) - \hat{j} \left(1 \cdot 1 - (-1) \cdot 3\right) + \hat{k} \left(1 \cdot (-4) - 2 \cdot 3\right) \]

\[ = \hat{i} (2 - 4) - \hat{j} (1 + 3) + \hat{k} (-4 - 6) \]

\[ = \hat{i} (-2) - \hat{j} (4) + \hat{k} (-10) \]

Thus, the normal vector \(n\) is:
\[ n = \langle -2, -4, -10 \rangle \]

### Step 3: Write the equation of the plane

The general equation of a plane can be expressed as:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((n_x, n_y, n_z)\) are the components of the normal vector.

Since the plane passes through the origin \((0, 0, 0)\), we can substitute \(x_0 = 0\), \(y_0 = 0\), and \(z_0 = 0\):
\[ -2(x - 0) - 4(y - 0) - 10(z - 0) = 0 \]

This simplifies to:
\[ -2x - 4y - 10z = 0 \]

Dividing through by -2 gives:
\[ x + 2y + 5z = 0 \]

### Final Answer

The equation of the required plane is:
\[ x + 2y + 5z = 0 \]